Single stanza followed by two stanza's of lyrics - lilypond

I'm looking for how to have a first set of lyrics with one stanza and following with more than one. Haven't been able to find an example. See how below the lyrics for first eight bars are unnecessarily repeated?
\version "2.18.2"
\header {
title = "Minimart"
subtitle = "Derived from Traditional Balkan Song: Rumelaj"
poet = "Text by Mike iLL"
}
melody = \relative c'' {
\clef treble
\key g \minor
\time 4/4
\repeat volta 2 { c4. b8 c b c( a) | r8 bes4 a8 bes( a) g r8 |
g a bes c d( c) bes c | g a4 a8 a4 a8 r8 | }
\repeat volta 2 { d8 cis d c d c d( c) | d cis d c d( c) bes( a) |
g bes b b b b c( b) | r a4 a8 bes( a) a8 r | }
}
text = \lyricmode {
There's a mi -- ni -- mart | on the cor -- ner |
At the mi -- ni -- mart is a | moon -- pie, I'm rea -- dy.
\set stanza = #"1. "
Need I need I need a high | Pen -- ny pen -- ny pen -- ny
Need I need I need a high | God al -- migh -- ty.
}
second_stanza = \lyricmode {
There's a mi -- ni -- mart | on the cor -- ner |
At the mi -- ni -- mart is a | moon -- pie, I'm rea -- dy.
\set stanza = #"2. "
Lift me lift me lift me high | Hea -- ven Hea -- ven Hea -- ven |
Lift me lift me lift me high | God al -- migh -- ty.
}
harmonies = \chordmode {
d1:7 | g1:m | g2:m/f g2:m/ees | d1:7
d1:7 | | g1:m | d1:7 |
}
\score {
<<
\new ChordNames {
\set chordChanges = ##t
\harmonies
}
\new Voice = "one" { \melody }
\new Lyrics \lyricsto "one" \text
\new Lyrics \lyricsto "one" \second_stanza
>>
\layout { }
\midi { }
}
If I remove the duplicate first half of the lyrics, the second half gets shifted to the beginning.

A solution to your problem is to use the following construction:
this is an example
<<
{this will be in the top}
\new Lyrics {and this in the bottom}
>>
only a single lyrics line once again from here on
In your particular case, this results in:
\version "2.18.2"
\header {
title = "Minimart"
subtitle = "Derived from Traditional Balkan Song: Rumelaj"
poet = "Text by Mike iLL"
}
melody = \relative c'' {
\clef treble
\key g \minor
\time 4/4
\repeat volta 2 { c4. b8 c b c( a) | r8 bes4 a8 bes( a) g r8 |
g a bes c d( c) bes c | g a4 a8 a4 a8 r8 | }
\repeat volta 2 { d8 cis d c d c d( c) | d cis d c d( c) bes( a) |
g bes b b b b c( b) | r a4 a8 bes( a) a8 r | }
}
text = \lyricmode {
There's a mi -- ni -- mart on the cor -- ner
At the mi -- ni -- mart is a moon -- pie, I'm rea -- dy.
<<
{
\set stanza = #"1. "
Need I need I need a high Pen -- ny pen -- ny pen -- ny
Need I need I need a high God al -- migh -- ty.
}
\new Lyrics {
\set associatedVoice = "melody"
\set stanza = #"2. "
Lift me lift me lift me high Hea -- ven Hea -- ven Hea -- ven
Lift me lift me lift me high God al -- migh -- ty.
}
>>
}
harmonies = \chordmode {
d1:7 | g1:m | g2:m/f g2:m/ees | d1:7
d1:7 | | g1:m | d1:7 |
}
\score {
<<
\new ChordNames {
\set chordChanges = ##t
\harmonies
}
\new Voice = "one" { \melody }
\new Lyrics \lyricsto "one" \text
>>
\layout { }
\midi { }
}

Related

How can add rightmost column by calculating all prices by id wise?

I have 4 tables. they are as follows:
-- 1. m_student
-- 2. t_reserve_a
-- 3. t_reserve_b
-- 4. t_reserve_c
I wanted to merge all required data from these 4 tables and get the total price of ordered textbooks in rightmost column as "total". i have tried to merge data from these table as follows:
SELECT tr.recptid AS ID,
tr.student_id AS S_ID,
tr.student_name AS S_N,
tr.student_zip AS ZIP,
tr.student_address AS ADD1,
tr.student_address2 AS ADD2,
tr.student_tel AS TEL,
IFNULL(tr.text_name, '') AS T_N,
tr.price,
COALESCE(tr.upd_date, rc.upd_date) AS UPDT
FROM
( SELECT ra.reserve_id,
RIGHT(ra.reserve_id,5) AS recptid,
m.student_id,
m.student_name,
m.student_zip,
m.student_address,
m.student_address2,
m.student_tel,
rb.subject,
rb.text_name,
rb.price,
rb.upd_date
FROM t_reserve_a AS ra INNER JOIN t_reserve_b AS rb ON ra.reserve_id=rb.reserve_id
INNER JOIN m_student AS m ON m.student_id=ra.student_id ) AS tr
LEFT JOIN t_reserve_c AS rc ON tr.reserve_id=rc.reserve_id
ORDER BY tr.reserve_id
they display like here in below:
ID S_ID S_N ZIP ADD1 ADD2 TEL T_N price UPDT
00000 19C2135004L H A 100-0000 USA TUCSON AZ 85705 0000-00-0000 GA 2475 2020/04/13 8:04:36
00000 19C2135004L H A 100-0000 USA TUCSON AZ 85705 0000-00-0000 R 1782 2020/04/13 8:04:36
00001 20A1109010F H S 100-0001 USA TUCSON AZ 706 0000-00-0001 R 1782 2020/04/13 8:05:51
00001 20A1109010F H S 100-0001 USA TUCSON AZ 706 0000-00-0001 LL 891 2020/04/13 8:05:51
00002 19C2134016F M K 100-0002 USA PHOENIX AZ 85123 0000-00-0002 R 2475 2020/04/13 8:06:27
00003 20A1124018B C O 100-0003 USA PHOENIX AZ 85122 0000-00-0003 LL 1782 2020/04/13 8:06:47
00003 20A1124018B C O 100-0003 USA PHOENIX AZ 85122 0000-00-0003 R 891 2020/04/13 8:06:47
00004 20C2136005I F H 100-0004 USA SEATTLE WA 98102 0000-00-0004 R 3534 2020/04/13 8:07:03
00004 20C2136005I F H 100-0004 USA SEATTLE WA 98102 0000-00-0004 T 2178 2020/04/13 8:07:03
00004 20C2136005I F H 100-0004 USA SEATTLE WA 98102 0000-00-0004 GB 4158 2020/04/13 8:07:03
00005 20C3156502C Y 100-0005 USA TUCSON AZ 8572 0000-00-0005 R 3534 2020/04/13 8:07:58
00005 20C3156502C Y 100-0005 USA TUCSON AZ 8572 0000-00-0005 GB 2178 2020/04/13 8:07:58
But how can I add total column like as below:
ID S_ID S_N ZIP ADD1 ADD2 TEL T_N price UPDT total
00000 19C2135004L H A 100-0000 USA TUCSON AZ 85705 0000-00-0000 GA 2475 2020/04/13 8:04:36 4257
00000 19C2135004L H A 100-0000 USA TUCSON AZ 85705 0000-00-0000 R 1782 2020/04/13 8:04:36
00001 20A1109010F H S 100-0001 USA TUCSON AZ 706 0000-00-0001 R 1782 2020/04/13 8:05:51 2673
00001 20A1109010F H S 100-0001 USA TUCSON AZ 706 0000-00-0001 LL 891 2020/04/13 8:05:51
00002 19C2134016F M K 100-0002 USA PHOENIX AZ 85123 0000-00-0002 R 2475 2020/04/13 8:06:27 2475
00003 20A1124018B C O 100-0003 USA PHOENIX AZ 85122 0000-00-0003 LL 1782 2020/04/13 8:06:47 2673
00003 20A1124018B C O 100-0003 USA PHOENIX AZ 85122 0000-00-0003 R 891 2020/04/13 8:06:47
00004 20C2136005I F H 100-0004 USA SEATTLE WA 98102 0000-00-0004 R 3534 2020/04/13 8:07:03 9870
00004 20C2136005I F H 100-0004 USA SEATTLE WA 98102 0000-00-0004 T 2178 2020/04/13 8:07:03
00004 20C2136005I F H 100-0004 USA SEATTLE WA 98102 0000-00-0004 GB 4158 2020/04/13 8:07:03
00005 20C3156502C Y 100-0005 USA TUCSON AZ 8572 0000-00-0005 R 3534 2020/04/13 8:07:58 5712
00005 20C3156502C Y 100-0005 USA TUCSON AZ 8572 0000-00-0005 GB 2178 2020/04/13 8:07:58
can anyone please help me to sort out this problem?

How to integrate midi export into existing lilypond score

I'm trying to do a MIDI export of existing Lilypond scores, but am unable to do so if there's more than one staff. I'm new to Lilypond but have been through the documentation and forums with no luck.
Wrapping the code below in \score doesn't work. If I take the first \relative c' section and delete everything after it, things appear to work fine, but I need the score in its entirety, including the paper and markup instructions.
Also tried \new Staff, as seen in some of the documentation, but ended up exactly where I started.
\score {
\header{
title = "Exercise: C, D, E Notes"
}
\paper {
#(set-paper-size "arch a" 'landscape)
system-system-spacing #'basic-distance = #20
markup-system-spacing #'basic-distance = #15
indent = 0\cm
}
\markup { \bold "Treble Clef - Right Hand" }
\markup { \small Fingering }
\relative c' {
\time 4/4
\override Staff.TimeSignature #'style = #'()
\clef "treble"
c4 c c c | d d d d | e e e e | e2 e | \break
e4 e e e | d d d d | c c c c | c2 c | \bar "|." \break
}
\markup { \bold "Bass Clef - Left Hand" }
\markup { \small Fingering }
\relative c, {
\time 4/4
\override Staff.TimeSignature #'style = #'()
\clef "bass"
e4 e e e | d d d d | c c c c | c2 c | \break
e4 e e e | d d d d | c c c c | c2 c | \bar "|."
}
\layout { }
\midi { }
}
\version "2.18.2" % necessary for upgrading to future LilyPond versions.
I would assign each of the little sections of music to individual variables and then create separate score blocks for each as well as a unique score block for midi output only:
\version "2.18.2" % necessary for upgrading to future LilyPond versions.
\header{
title = "Exercise: C, D, E Notes"
}
\paper{
#(set-paper-size "arch a" 'landscape)
system-system-spacing.basic-distance = #20
markup-system-spacing.basic-distance = #15
indent = 0\cm
}
mark_A = ^\markup { \bold "Treble Clef - Right Hand" }^\markup { \small Fingering }
mark_B = ^\markup { \bold "Bass Clef - Left Hand" }^\markup { \small Fingering }
music_A = \relative c' {
\time 4/4
\override Staff.TimeSignature #'style = #'()
\clef "treble"
c4-\mark_A c c c | d d d d | e e e e | e2 e | \break
e4 e e e | d d d d | c c c c | c2 c | \bar "|." \break
}
music_B = \relative c, {
\time 4/4
\override Staff.TimeSignature #'style = #'()
\clef "bass"
e4-\mark_B e e e | d d d d | c c c c | c2 c | \break
e4 e e e | d d d d | c c c c | c2 c | \bar "|."
}
\score{
\new Staff \music_A
\layout{}
}
\score{
\new Staff \music_B
\layout{}
}
\score{
\new Staff {\music_A \music_B}
\midi{}
}
More info here: http://lilypond.org/doc/v2.19/Documentation/learning/multiple-staves

Stacking columns without melting by id

I imported a json file using rjson and converted it to a data.frame but all of the data is spread widthwise and the column names contain the key information.
stations <- fromJSON(file = "station_information.json")
test <- as.data.frame(stations[3])
What this looks like is:
> dim(test)
[1] 2 5985
> test[1:27]
data.stations.station_id data.stations.name data.stations.short_name
1 72 W 52 St & 11 Ave 6926.01
2 72 W 52 St & 11 Ave 6926.01
data.stations.lat data.stations.lon data.stations.region_id
1 40.76727 -73.99393 71
2 40.76727 -73.99393 71
data.stations.rental_methods data.stations.capacity
1 KEY 39
2 CREDITCARD 39
data.stations.eightd_has_key_dispenser data.stations.station_id.1
1 FALSE 79
2 FALSE 79
data.stations.name.1 data.stations.short_name.1 data.stations.lat.1
1 Franklin St & W Broadway 5430.08 40.71912
2 Franklin St & W Broadway 5430.08 40.71912
data.stations.lon.1 data.stations.region_id.1 data.stations.rental_methods.1
1 -74.00667 71 KEY
2 -74.00667 71 CREDITCARD
data.stations.capacity.1 data.stations.eightd_has_key_dispenser.1
1 33 FALSE
2 33 FALSE
data.stations.station_id.2 data.stations.name.2 data.stations.short_name.2
1 82 St James Pl & Pearl St 5167.06
2 82 St James Pl & Pearl St 5167.06
data.stations.lat.2 data.stations.lon.2 data.stations.region_id.2
1 40.71117 -74.00017 71
2 40.71117 -74.00017 71
data.stations.rental_methods.2 data.stations.capacity.2
1 KEY 27
2 CREDITCARD 27
data.stations.eightd_has_key_dispenser.2
1 FALSE
2 FALSE
As you can see, this cannot be fixed with a simple transpose t() or melt() solution. I am wondering what I am doing wrong on import or conversion to data.frame which leaves me with a dataframe stretched with the index of what should be rows appended to the column names.
I have tried both of these methods but I am left with the same stretched data:
plyr::ldply(stations, data.frame)
do.call(rbind, lapply(stations, data.frame, stringsAsFactors=FALSE))
In the end I would like my output to look like every 9 columns were "cut" and stacked on to the first 9 - so that I am left with 655 rows and 9 columns Any suggestions would be appreciated.
NOTE: I am taking the JSON directly from this link (it is not a large file)
Here is a reproducible example of the first 27 columns, which should be reshaped to a 9 x 3 dataframe:
> dput(df)
structure(list(data.stations.station_id = structure(c(1L, 1L), class = "factor", .Label = "72"),
data.stations.name = structure(c(1L, 1L), class = "factor", .Label = "W 52 St & 11 Ave"),
data.stations.short_name = structure(c(1L, 1L), class = "factor", .Label = "6926.01"),
data.stations.lat = c(40.76727216, 40.76727216), data.stations.lon = c(-73.99392888,
-73.99392888), data.stations.region_id = c(71, 71), data.stations.rental_methods = structure(c(2L,
1L), .Label = c("CREDITCARD", "KEY"), class = "factor"),
data.stations.capacity = c(39, 39), data.stations.eightd_has_key_dispenser = c(FALSE,
FALSE), data.stations.station_id.1 = structure(c(1L, 1L), class = "factor", .Label = "79"),
data.stations.name.1 = structure(c(1L, 1L), class = "factor", .Label = "Franklin St & W Broadway"),
data.stations.short_name.1 = structure(c(1L, 1L), class = "factor", .Label = "5430.08"),
data.stations.lat.1 = c(40.71911552, 40.71911552), data.stations.lon.1 = c(-74.00666661,
-74.00666661), data.stations.region_id.1 = c(71, 71), data.stations.rental_methods.1 = structure(c(2L,
1L), .Label = c("CREDITCARD", "KEY"), class = "factor"),
data.stations.capacity.1 = c(33, 33), data.stations.eightd_has_key_dispenser.1 = c(FALSE,
FALSE), data.stations.station_id.2 = structure(c(1L, 1L), class = "factor", .Label = "82"),
data.stations.name.2 = structure(c(1L, 1L), class = "factor", .Label = "St James Pl & Pearl St"),
data.stations.short_name.2 = structure(c(1L, 1L), class = "factor", .Label = "5167.06"),
data.stations.lat.2 = c(40.71117416, 40.71117416), data.stations.lon.2 = c(-74.00016545,
-74.00016545), data.stations.region_id.2 = c(71, 71), data.stations.rental_methods.2 = structure(c(2L,
1L), .Label = c("CREDITCARD", "KEY"), class = "factor"),
data.stations.capacity.2 = c(27, 27), data.stations.eightd_has_key_dispenser.2 = c(FALSE,
FALSE)), .Names = c("data.stations.station_id", "data.stations.name",
"data.stations.short_name", "data.stations.lat", "data.stations.lon",
"data.stations.region_id", "data.stations.rental_methods", "data.stations.capacity",
"data.stations.eightd_has_key_dispenser", "data.stations.station_id.1",
"data.stations.name.1", "data.stations.short_name.1", "data.stations.lat.1",
"data.stations.lon.1", "data.stations.region_id.1", "data.stations.rental_methods.1",
"data.stations.capacity.1", "data.stations.eightd_has_key_dispenser.1",
"data.stations.station_id.2", "data.stations.name.2", "data.stations.short_name.2",
"data.stations.lat.2", "data.stations.lon.2", "data.stations.region_id.2",
"data.stations.rental_methods.2", "data.stations.capacity.2",
"data.stations.eightd_has_key_dispenser.2"), row.names = c(NA,
-2L), class = "data.frame")
So the output structure should look like this (obviously with the values not NA). Each row represents the appended index number of the original dataframe's column names
> output
data.stations.station_id data.stations.name data.stations.short_name
1 NA NA NA
2 NA NA NA
3 NA NA NA
data.stations.lat data.stations.lon data.stations.region_id
1 NA NA NA
2 NA NA NA
3 NA NA NA
data.stations.rental_methods data.stations.capacity
1 NA NA
2 NA NA
3 NA NA
data.stations.eightd_has_key_dispenser
1 NA
2 NA
3 NA
I would try:
library(data.table)
rbindlist(lapply(split(seq_along(df), c(0, (seq_along(df)%/%9)[-length(df)])),
function(x) df[, x]), use.names = FALSE)
## data.stations.station_id data.stations.name data.stations.short_name data.stations.lat
## 1: 72 W 52 St & 11 Ave 6926.01 40.76727
## 2: 72 W 52 St & 11 Ave 6926.01 40.76727
## 3: 79 Franklin St & W Broadway 5430.08 40.71912
## 4: 79 Franklin St & W Broadway 5430.08 40.71912
## 5: 82 St James Pl & Pearl St 5167.06 40.71117
## 6: 82 St James Pl & Pearl St 5167.06 40.71117
## data.stations.lon data.stations.region_id data.stations.rental_methods
## 1: -73.99393 71 KEY
## 2: -73.99393 71 CREDITCARD
## 3: -74.00667 71 KEY
## 4: -74.00667 71 CREDITCARD
## 5: -74.00017 71 KEY
## 6: -74.00017 71 CREDITCARD
## data.stations.capacity data.stations.eightd_has_key_dispenser
## 1: 39 FALSE
## 2: 39 FALSE
## 3: 33 FALSE
## 4: 33 FALSE
## 5: 27 FALSE
## 6: 27 FALSE
That is, create a list of data.frames with 9 columns each, and rbind them. This way, you won't have problems of data coercion when converting to a matrix.
This results in a 6 row x 9 column data.table. Not sure what rule you want to use to drop rows to end up with just 3 rows....
But I think you're trying to solve a problem that does not exist. Try reading your data like this:
library(jsonlite)
x <- fromJSON("https://gbfs.citibikenyc.com/gbfs/en/station_information.json")
head(x[[3]]$stations)
## station_id name short_name lat lon region_id
## 1 72 W 52 St & 11 Ave 6926.01 40.76727 -73.99393 71
## 2 79 Franklin St & W Broadway 5430.08 40.71912 -74.00667 71
## 3 82 St James Pl & Pearl St 5167.06 40.71117 -74.00017 71
## 4 83 Atlantic Ave & Fort Greene Pl 4354.07 40.68383 -73.97632 71
## 5 116 W 17 St & 8 Ave 6148.02 40.74178 -74.00150 71
## 6 119 Park Ave & St Edwards St 4700.06 40.69609 -73.97803 71
## rental_methods capacity eightd_has_key_dispenser
## 1 KEY, CREDITCARD 39 FALSE
## 2 KEY, CREDITCARD 33 FALSE
## 3 KEY, CREDITCARD 27 FALSE
## 4 KEY, CREDITCARD 62 FALSE
## 5 KEY, CREDITCARD 39 FALSE
## 6 KEY, CREDITCARD 19 FALSE
dim(x[[3]]$stations)
# [1] 665 9
You can use matrix but make sure all your factor columns are characters, i.e.
ind <- sapply(df, is.factor)
df[ind] <- lapply(df[ind], as.character)
final_df <- as.data.frame(matrix(unlist(df), ncol = 9, byrow = TRUE))
final_df[c(TRUE, FALSE),]
# V1 V2 V3 V4 V5 V6 V7 V8 V9
#1 72 72 W 52 St & 11 Ave W 52 St & 11 Ave 6926.01 6926.01 40.76727216 40.76727216 -73.99392888
#3 79 79 Franklin St & W Broadway Franklin St & W Broadway 5430.08 5430.08 40.71911552 40.71911552 -74.00666661
#5 82 82 St James Pl & Pearl St St James Pl & Pearl St 5167.06 5167.06 40.71117416 40.71117416 -74.00016545
On the other hand, as #A5C1D2H2I1M1N2O1R2T1 notes, you might be looking for this instead:
as.data.frame(matrix(c(t(df)), ncol = 9, byrow = TRUE))
# V1 V2 V3 V4 V5 V6 V7 V8 V9
#1 72 W 52 St & 11 Ave 6926.01 40.76727 -73.99393 71 KEY 39 FALSE
#2 79 Franklin St & W Broadway 5430.08 40.71912 -74.00667 71 KEY 33 FALSE
#3 82 St James Pl & Pearl St 5167.06 40.71117 -74.00017 71 KEY 27 FALSE
#4 72 W 52 St & 11 Ave 6926.01 40.76727 -73.99393 71 CREDITCARD 39 FALSE
#5 79 Franklin St & W Broadway 5430.08 40.71912 -74.00667 71 CREDITCARD 33 FALSE
#6 82 St James Pl & Pearl St 5167.06 40.71117 -74.00017 71 CREDITCARD 27 FALSE

how to get two tables data based on column

We have two tables Table1 and table2. Table1 have A,B,C columns and Table2 have B,E,F columns.
Table1 Table2
A B C B E F
raja 21 HYD 21 AP 60000000
raju 21 VJY
ravi 21 BEG
We need is
A B C E F
raja 21 HYD AP 60000000
raju 21 VJY AP 60000000
ravi 21 BEG AP 60000000
so We tried like this
tx.executeSql('SELECT r.A as name,r.B as age,r.C as city,s.E as state,s.F as distance from table1 r Left JOIN table2 s on r.B=s.B',[],joinsuccess,errorCB);
function joinsuccess(tx,result){
var JOINLength = result.rows.length;
for (var i=0; i<JOINLength; i++){
console.log(result.rows.item(i).name);
console.log(result.rows.item(i).age);
console.log(result.rows.item(i).city);
console.log(result.rows.item(i).state);
console.log(result.rows.item(i).distance);
}
}
but it's we got only
ravi 21 BEG AP 60000000

Code Golf - Banner Generation

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
When thanking someone, you don't want to just send them an e-mail saying "Thanks!", you want to have something FLASHY:
Input: THANKS!!
Output:
TTT H H AAA N N K K SSS !!! !!!
T H H A A NNN K K S !!! !!!
T HHH AAA NNN KK SSS !!! !!!
T H H A A N N K K S
T H H A A N N K K SSS !!! !!!
Write a program to generate a banner. You only have to generate upper-case A-Z along with spaces and exclamation points (what is a banner without an exclamation point?). All characters are made up of a 3x5 grid of the same character (so the S is a 3x5 grid made of S). All output should be on one row (so no newlines). Here are all the letters you need:
Input: ABCDEFGHIJKL
Output:
AAA BBB CCC DD EEE FFF GGG H H III JJJ K K L
A A B B C D D E F G H H I J K K L
AAA BBB C D D EE FF G G HHH I J KK L
A A B B C D D E F G G H H I J J K K L
A A BBB CCC DD EEE F GGG H H III JJJ K K LLL
Input: MNOPQRSTUVWX
Output:
M M N N OOO PPP QQQ RR SSS TTT U U V V W W X X
MMM NNN O O P P Q Q R R S T U U V V W W X
M M NNN O O PPP Q Q RR SSS T U U V V WWW X
M M N N O O P QQQ R R S T U U V V WWW X
M M N N OOO P QQQ R R SSS T UUU V WWW X X
Input: YZ!
Output:
Y Y ZZZ !!!
Y Y Z !!!
YYY Z !!!
Y Z
YYY ZZZ !!!
The winner is the shortest source code, as counted by the number of bytes it takes to store the file in utf-8 encoding. Source code should read input from stdin, output to stdout. You can assume input will only contain [A-Z! ]. If you insult the user on incorrect input, you get a 10 character discount =P.
I was going to require these exact 28 characters, but to make it more interesting, you can choose how you want them to look - whatever makes your code shorter! To prove that your letters do look like normal letters, show the output of the last three runs.
Shortest codes so far, in characters (utf8 encoding if non-ASCII present):
133 J
205 Python
209 Ruby
313 Haskell
345 C89
382 F#
J, 133 135 79 83 84 88 characters (utf-8 encoding)
;/5 3$"1(' ',.s){~"1#:3 u:(ucp'翇篭篯礧歮禧禤祯寭璗牯宭䤧彭忭筯篤筿殭秏璒孯孪寿咕寏犧'){~0>.64-~a.i.s=:
Usage:
;/5 3$"1(' ',.s){~"1#:3 u:(ucp'翇篭篯礧歮禧禤祯寭璗牯宭䤧彭忭筯篤筿殭秏璒孯孪寿咕寏犧'){~0>.64-~a.i.s=:'ABCDEFGHIJKLMNOPQRSTUVWXYZ !'
┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐
│AAA│BBB│CCC│DD │EEE│FFF│GGG│H H│III│JJJ│K K│L │M M│N N│OOO│PPP│QQQ│RR │SSS│TTT│U U│V V│W W│X X│Y Y│ZZZ│ │!!!│
│A A│B B│C │D D│E │F │G │H H│ I │ J│K K│L │MMM│NNN│O O│P P│Q Q│R R│S │ T │U U│V V│W W│ X │Y Y│ Z│ │!!!│
│AAA│BBB│C │D D│EE │FF │G G│HHH│ I │ J│KK │L │M M│NNN│O O│PPP│Q Q│RR │SSS│ T │U U│V V│WWW│ X │YYY│ Z │ │!!!│
│A A│B B│C │D D│E │F │G G│H H│ I │J J│K K│L │M M│N N│O O│P │QQQ│R R│ S│ T │U U│V V│WWW│ X │ Y│Z │ │ │
│A A│BBB│CCC│DD │EEE│F │GGG│H H│III│JJJ│K K│LLL│M M│N N│OOO│P │QQQ│R R│SSS│ T │UUU│ V │WWW│X X│YYY│ZZZ│ │!!!│
└───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┘
;/5 3$"1(' ',.s){~"1#:3 u:(ucp'翇篭篯礧歮禧禤祯寭璗牯宭䤧彭忭筯篤筿殭秏璒孯孪寿咕寏犧'){~0>.64-~a.i.s=:'this is incorrect input.'
|index error
Explanation (NB. is comment in J):
;/ NB. String together along the third dimension...
5 3$"1 NB. ... reshape each line to 5x3...
(' ',.s) NB. ... a space before each letter of the input string...
{~"1 NB. ... indexed using...
#: NB. ... the (15 bit) binary representation of ...
3 u: NB. ... the integer representation of...
(ucp'翇篭篯礧歮禧禤祯寭璗牯宭䤧彭忭筯篤筿殭秏璒孯孪寿咕寏犧') ... the unicode versions of these code points...
{~ NB. ...indexed using...
0>. NB. ...the max of 0 and...
64-~ NB. ...64 less than...
a.i. NB. the ascii indexes of s
s=: NB. Assign the input string to the variable s.
Python, 250 224 chars
s=raw_input()
for i in range(5):
for c in s:
print''.join((' ',c)[int('2zj93fqzj6hsh2bc8i2b1ycncj5yc2v9i0m16dz91gcizj18blbw6wt0p3qqh8svchwc5onna2808of',36)>>((ord(c)-65 if c>'#'else 26)*15+i*3+j)&1]for j in[0,1,2]),
print
Notes:
relies on 2.x print statement considerably;
supports spaces.
Running (I changed appearance of a few letters, for aestetic reasons only ;):
$ echo ABCDEFGHIJKL | python code-golf.py
AAA BBB CCC DD EEE FFF GGG H H III JJJ K K L
A A B B C D D E F G H H I J K K L
AAA BBB C D D EEE FFF G HHH I J KK L
A A B B C D D E F G G H H I J J K K L
A A BBB CCC DD EEE F GGG H H III JJJ K K LLL
$ echo MNOPQRSTUVWX | python code-golf.py
M M N N OOO PPP QQQ RR SSS TTT U U V V W W X X
MMM NNN O O P P Q Q R R S T U U V V W W X X
M M NNN O O PPP Q Q RR SSS T U U V V WWW X
M M N N O O P QQQ R R S T U U V V WWW X X
M M N N OOO P QQQ R R SSS T UUU V W W X X
$ echo YZ\! | python code-golf.py
Y Y ZZZ !!!
Y Y Z !!!
YYY Z !!!
Y Z
YYY ZZZ !!!
There are trailing spaces.
Figlet (0 chars)
wget -q 3.ly/gzkv;figlet -f b
Sample output:
% wget -q 3.ly/gzkv;figlet -f b ABCDEFGHIJKLMNOPQRS
A BB CC DD EEE FFF GG H H III JJ K K L M M NNN O PP Q RR SS
A A B B C D D E F G H H I J K K L MMM N N O O P P Q Q R R S
AAA BB C D D EE FF G G HHH I J KK L MMM N N O O PP Q Q RR S
A A B B C D D E F G G H H I J J K K L M M N N O O P QQ R R S
A A BB CC DD EEE F GG H H III J K K LLL M M N N O P Q R R SS
% wget -q 3.ly/gzkv;figlet -f b TUVWXYZ \!
TTT U U V V W W X X Y Y ZZZ !!!
T U U V V W W X X Y Y Z !!!
T U U V V WWW X Y Z !!!
T U U V V WWW X X Y Z
T UUU V W W X X Y ZZZ !!!
Python, 233 231 225 224 205 chars
Candidate for the shortest python solution here :-)
It is a two-liner - last line was broken in 3 for urrrr readability
s=raw_input()
for i in range(5):print' '.join(''.join((' ',c)[ord(
'W_E__U__QQ_QN_UQ_EA_Q]_D_Q_QYQ__D[_PP_B__F__Q__EG_Y__EZWU]A_A_P_OPO_\\_QNQWT_YUS'
[max(0,3*ord(c)-192-k)])>>i&1]for k in(2,1,0))for c in s)
ps. Thanks for comments, corrected issues and now using 7-bit ASCII only!
Test banner:
TTT H H EEE QQQ U U III CCC K K BBB RR OOO W W N N FFF OOO X X JJJ U U M M PPP SSS OOO V V EEE RR TTT H H EEE L AAA ZZZ Y Y DD OOO GGG !!!
T H H E Q Q U U I C K K B B R R O O W W NNN F O O X J U U MMM P P S O O V V E R R T H H E L A A Z Y Y D D O O G !!!
T HHH EE Q Q U U I C KK BBB RR O O WWW NNN FF O O X J U U M M PPP SSS O O V V EE RR T HHH EE L AAA Z YYY D D O O G G !!!
T H H E QQQ U U I C K K B B R R O O WWW N N F O O X J J U U M M P S O O V V E R R T H H E L A A Z Y D D O O G G
T H H EEE QQQ UUU III CCC K K BBB R R OOO WWW N N F OOO X X JJJ UUU M M P SSS OOO V EEE R R T H H EEE LLL A A ZZZ YYY DD OOO GGG !!!
Haskell, 313 316 320
import Data.Bits
import Data.Char
c&True=c
c&_=' '
a ' '='#'
a '!'='['
a c=c
q s=unlines[s>>= \c->take 3(drop(84*n+3*(ord(a c)-64))$map((c&).testBit(0xffdebaf79f6fbfde7bfe8062f6a979b69b55a4d368ebaf6aeefbe9717add3f8f2ab6a36dbf9b1524d368fedb6fefff69bfdffbff8::Integer))[0..])++" "|n<-[0..4]]
main=getLine>>=putStr.q
For the curious, the large number is the hex version of encoding in the following code. The number is simply used as a bitmap. I've had no success with further shortening the code by encoding the number to other bases even with non-standard character representations.
formats :: [String] -- order: [ A-Z!] <- that's a space in front of A
formats = [
" AAABBBCCCDD EEEFFFGGGH HIIIJJJK KL M MN NOOOPPPQQQRR SSSTTTU UV VW WX XY YZZZ!!!"
, " A AB BC D DE F G H H I JK KL MMMNNNO OP PQ QR RS T U UV VW W X Y Y Z!!!"
, " AAABBBC D DEE FF G GHHH I JKK L M MNNNO OPPPQ QRR SSS T U UV VWWW X YYY Z !!!"
, " A AB BC D DE F G GH H I J JK KL M MN NO OP QQQR R S T U UV VWWW X YZ "
, " A ABBBCCCDD EEEF GGGH HIIIJJJK KLLLM MN NOOOP QQQR RSSS T UUU V WWWX XYYYZZZ!!!"
]
charToBool :: Char -> Bool
charToBool ' ' = False
charToBool _ = True
boolToInteger :: Bool -> Integer
boolToInteger True = 1
boolToInteger _ = 0
encoding :: Integer
encoding = foldr f 0 $ zip [0..] $ map charToBool $ concat formats
where
f (pow, bool) z = z + ((2^pow) * boolToInteger bool)
Python 2.6, 251 - 243 - 227 characters
I tried a slightly different approach (bitpacking the parts the letters are made of) ...
handles uppercase letters, spaces, exclamation mark.
2 unnecessary linebreaks added here for readability (the for loop could be one line)
be sure to save this as UTF-8 with BOM!
As always, all comments and suggestions welcome! Contender for shortest Python solution (again a few characters behind at the moment)...
w=raw_input()
for l in range(5):print''.join("1111 11 11 1 1 "
[int(("%05d"%ord(u"<ϳϲࢬ禉ऐऒ࠾⬃ᅘᖆⰯ囌❿✛іϾь穏ࠂᅜ⭦⭪⫸㡩⬪㰼"
[max(0,ord(c)-64)]))[l])*3:][:3].replace("1",c)+" "for c in w)
C89, 345 characters
Newlines added for your sanity (they are not included in the character count and can/should be deleted):
char o[5][99];
d[]={0x2df7fbef,0x3927bb6b,0x396792cf,0x3da7dbed,0x3a4bfb27,0x2d76f249,0x2dbedbfd,0x3db793ef,0x3fb7daeb,0x3ce7a497,0x3db6ab6d,0x3ff6d495,0x3cf6f2a7,0x38ff8000};
c,i,j;
main(){memset(o,32,495);
while((c=getchar())>0){
for(j=0;j<15;j++)
o[j/3][i+j%3]=d[c-33?(c-65)/2:13]>>((c&1)*15+j)&1?c:32;i+=4;}
for(j=0;j<5;j++)printf("%.*s\n",i,o[j]);}
Ruby : 207 215 252 345 characters
i=gets.chomp;5.times{|t|p i.gsub(/./){|c|j=3*(c>?#?c.ord-64:0);(3*t..3*t+2).map{|d|"mini5mbmzjf2bqjmof3prl72i5pn138iuhylmkpi65i278kq3qjfaihyjb66787odp8ktiy5hwt78tmnb"[j..j+2].to_i(36)[d]==1?c:" "}.join+" "}}
F#, 382 chars
I compressed two letters from each row into an ascii-printable byte, and handled space and exclamation point specially.
let s,(!)=stdin.ReadLine(),printf"%s"
for n in 0..4 do
for c in s do if c=' '||n=3&&c='!'then !" "elif c='!'then !"!!! "else for x in 0..3 do printf"%c"(if(Array.collect(fun b->let B n=int b&&&n=0 in[|(B 64)||not(B 8);B 32;B 16;true;B 4;B 2;B 1;true|])"wvwuwTUwvwUUWUEDEiTwUUBURQwEfWidWWVrUrrUEDUmTUTuZUr\\WvtuwWUturruw"B).[n*104+(int c-int 'A')*4+x]then ' 'else c)
!"\n"
Sample I/O below:
HELLO WORLD!!!
H H EEE L L OOO W W OOO RR L DD !!! !!! !!!
H H E L L O O W W O O R R L D D !!! !!! !!!
HHH EE L L O O WWW O O RR L D D !!! !!! !!!
H H E L L O O WWW O O R R L D D
H H EEE LLL LLL OOO WWW OOO R R LLL DD !!! !!! !!!
ABCDEFGHIJKL
AAA BBB CCC DD EEE FFF GGG H H III JJJ K K L
A A B B C D D E F G H H I J K K L
AAA BBB C D D EE FF G G HHH I J KK L
A A B B C D D E F G G H H I J J K K L
A A BBB CCC DD EEE F GGG H H III JJJ K K LLL
MNOPQRSTUVWXYZ
M M N N OOO PPP QQQ RR SSS TTT U U V V W W X X Y Y ZZZ
MMM NNN O O P P Q Q R R S T U U V V W W X Y Y Z
M M NNN O O PPP Q Q RR SSS T U U V V WWW X YYY Z
M M N N O O P QQQ R R S T U U V V WWW X Y Z
M M N N OOO P QQQ R R SSS T UUU V WWW X X YYY ZZZ
Python, 340 characters
d=dict((i,[23535,31727,29263,15211,29391,4815,31567,23533,29847,31527,23277,29257,23421,23549,31599,5103,32623,23275,31183,9367,31597,11117,32749,21653,31213,29351][i-65])for i in range(65,91))
d[33]=29183
d[32]=0
s=raw_input()
for l in range(5):
p=""
for c in s:
for n in range(3):
if d[ord(c)]&2**(3*l+n):p+=c
else:p+=" "
p+=" "
print p
sample output
>>>
ABCDEFGHIJKLMNOPQRSTUVWXYZ !
aaa bbb ccc dd eee fff ggg h h iii jjj k k l m m n n ooo ppp qqq rr sss ttt u u v v w w x x y y zzz !!!
a a b b c d d e f g h h i j k k l mmm nnn o o p p q q r r s t u u v v w w x y y z !!!
aaa bbb c d d ee ff g g hhh i j kk l m m nnn o o ppp q q rr sss t u u v v www x yyy z !!!
a a b b c d d e f g g h h i j j k k l m m n n o o p qqq r r s t u u v v www x y z
a a bbb ccc dd eee f ggg h h iii jjj k k lll m m n n ooo p qqq r r sss t uuu v www x x yyy zzz !!!
>>>
not too great, but it was fun writing it
edit whoops, I made the input be lowercase. fixed now, saved me one character too :)
Delphi, 397 chars
Ok, with all the begin/end statements Delphi probably will never be shorter than any other languages, but I do see a challenge in getting it as short as possible.
vvar s:String;i,j,k:Word;const F:Array[65..92]of Word=($5BEA,$3AEB,$624E,$3B6B,$72CF,$12CF,$6B4E,$5BED,$7497,$2B26,$5AED,$7249,$5BFD,$5B6F,$2B6A,$12EB,$4D6A,$5AEB,$388E,$2497,$7B6D,$2B6D,$5FED,$5AAD,$24AD,$72A7,$2092,$0000);begin S:=ParamStr(1);for j:=0 to 4 do begin for k:=1 to Length(S)do begin for i := 0 to 2 do Write((' '+S[k])[1+(F[ord(S[k])]shr(i+j*3))and 1]);Write(' ');end;WriteLn;end;end.
The font is built up like this:
010 110 011 110 111 111 011 101 111 011 101 100 101 111 010 110 010 110 011 111 101 101 101 101 101 111 010 000
101 101 100 101 100 100 100 101 010 001 101 100 111 101 101 101 101 101 100 010 101 101 101 101 101 001 010 000
111 110 100 101 110 110 101 111 010 001 110 100 111 101 101 110 101 110 010 010 101 101 111 010 010 010 010 000
101 101 100 101 100 100 101 101 010 101 101 100 101 101 101 100 011 101 001 010 101 101 111 101 010 100 000 000
101 110 011 110 111 100 011 101 111 010 101 111 101 101 010 100 001 101 110 010 111 010 101 101 010 111 010 000
The characters in this 5x3 font take up 15 bits and are stored in word (UINT16) in this order:
00 01 02
03 04 05
06 07 08
09 10 11
12 13 14
Formatted code:
var
s:String;
i, j, k: Word;
const
F: Array [65 .. 92] of Word = (
$5BEA,$3AEB,$624E,$3B6B, $72CF,$12CF,$6B4E,$5BED,
$7497,$2B26,$5AED,$7249, $5BFD,$5B6F,$2B6A,$12EB,
$4D6A,$5AEB,$388E,$2497, $7B6D,$2B6D,$5FED,$5AAD,
$24AD,$72A7,$2092,$0000);
begin
S := ParamStr(1);
for j := 0 to 4 do
begin
for k := 1 to Length(S) do
begin
for i := 0 to 2 do
Write((' '+S[k])[1+(F[ord(S[k])]shr(i+j*3))and 1]);
Write(' ');
end;
WriteLn;
end;
end.
Python: 259 chars
Not the shortest, but considering it was my first Python script, I'm more than satisfied.
k=raw_input()
for i in range(5):print' '.join(''.join((' ',x)[int(z)]for z in bin(int(''.join('%02d'%(ord(q)-43)for q in'xwxvxabxwxbbdqbXWX#axbbUb_^qxXwd#kddcsbssqbXWvDabav7bs9+dwuvxdbuvssvxq')[i*28+'ABCDEFGHIJKLMNOPQRSTUVWXYZ! '.find(x)],8))[2:])for x in k)
Perl, 69 74 77 78 79 chars
$a=<>;s:(?{$z=substr$a,$-[0]/4,1})z|#:$z:g,print for`figlet -f3x5 $a`
Sample output:
% echo ABCDEFGHIJKLMNOPQRS | perl banner.pl
A BB CC DD EEE FFF GG H H III JJ K K L M M NNN O PP Q RR SS
A A B B C D D E F G H H I J K K L MMM N N O O P P Q Q R R S
AAA BB C D D EE FF G G HHH I J KK L MMM N N O O PP Q Q RR S
A A B B C D D E F G G H H I J J K K L M M N N O O P QQ R R S
A A BB CC DD EEE F GG H H III J K K LLL M M N N O P Q R R SS
% echo TUVWXYZ \! | perl banner.pl
TTT U U V V W W X X Y Y ZZZ !
T U U V V W W X X Y Y Z !
T U U V V WWW X Y Z !
T U U V V WWW X X Y Z
T UUU V W W X X Y ZZZ !
I assume you have figlet and this figlet font installed on your system. :)
C#, 239 231 229 chars (292 bytes)
I'm a bit late but this just looked like fun.
using C=System.Console;class P{static void Main(){var t=C.ReadLine();for(int
b=15,s;b>0;b-=3){foreach(var c in t)for(s=0;s++<4;)C.Write(s>3||c<33?' ':((
"翇篭篯礧歮禧禤祯寭璗牯宭䤧彭忭筯篤筿殭秏璒孯孪寿咕寏犧"[c<34?0:c-64])&1<<b-s)>0
?c:' ');C.WriteLine();}}}