I wonder how to find the position of one country in CoronaVirus update list https://www.worldometers.info/coronavirus by using Chrome console
I tried: $x("count(.//*[#id='main_table_countries_today']/tbody/tr/td[1][.='Vietnam']/preceding-sibling::*)")
Anyone could give me a light here. Thanks.
In your xpath you are applying preciding-siblings to the first td in a tr, but you should count siblings for tr as those are actually the rows.
This one worked for me:
$x("count(//table[#id='main_table_countries_today']//tr[td/text() = 'Vietnam']/preceding-sibling::*)")
Also in my opinion there is no need to do +1 at the end as the first row is irrelevant: the whole world. But you should know better, what you need.
Or here is yours fixed:
$x("count(.//table[#id='main_table_countries_today']/tbody/tr[td[1][.='Vietnam']]/preceding-sibling::*)")
To complete kishkin's answer, shortest form :
count((//td[.='Vietnam'])[1]/preceding::tr[#style=""])+1
Output : 93 (92+1)
And to exclude the Diamond Princess :
count((//td[.='Vietnam'])[1]/preceding::tr[#style=""][not(.//span)])+1
Output : 92 (91+1)
Related
I want to compare two smoothing methods for a bigram model:
Add-one smoothing
Interpolated Absolute Discounting
For the first method, I found some codes.
def calculate_bigram_probabilty(self, previous_word, word):
bigram_word_probability_numerator = self.bigram_frequencies.get((previous_word, word), 0)
bigram_word_probability_denominator = self.unigram_frequencies.get(previous_word, 0)
if self.smoothing:
bigram_word_probability_numerator += 1
bigram_word_probability_denominator += self.unique__bigram_words
return 0.0 if bigram_word_probability_numerator == 0 or bigram_word_probability_denominator == 0 else float(
bigram_word_probability_numerator) / float(bigram_word_probability_denominator)
However, I found nothing for the second method except for some references for 'KneserNeyProbDist'. However, this is for trigrams!
How can I change my code above to calculate it? The parameters of this method must be estimated from a development-set.
In this answer I just clear up a few things that I just found about your problem, but I can't provide a coded solution.
with KneserNeyProbDist you seem to refer to a python implementation of that problem: https://kite.com/python/docs/nltk.probability.KneserNeyProbDist
There exists an article about Kneser–Ney smoothing on wikipedia: https://en.wikipedia.org/wiki/Kneser%E2%80%93Ney_smoothing
The article above links this tutorial: https://nlp.stanford.edu/~wcmac/papers/20050421-smoothing-tutorial.pdf but this has a small fault on the most important page 29, the clear text is this:
Modified Kneser-Ney
Chen and Goodman introduced modified Kneser-Ney:
Interpolation is used instead of backoff. Uses a separate discount for one- and two-counts instead of a single discount for all counts. Estimates discounts on held-out data instead of using a formula
based on training counts.
Experiments show all three modifications improve performance.
Modified Kneser-Ney consistently had best performance.
Regrettable the modified Version is not explained in that document.
The original documentation by Chen & Goodman luckily is available, the Modified Kneser–Ney smoothing is explained on page 370 of this document: http://u.cs.biu.ac.il/~yogo/courses/mt2013/papers/chen-goodman-99.pdf.
I copy the most important text and formula here as screenshot:
So the Modified Kneser–Ney smoothing now is known and seems being the best solution, just translating the description beside formula in running code is still one step to do.
It might be helpful that below the shown text (above in screenshot) in the original linked document is still some explanation that might help to understand the raw description.
I have the object TABLE_LIST which is a list that has tables (I can't provide the contents for privacy policies, sorry).
I first created the object TABLE_LIST (It is a list of data.frames 2x12)
TABLE_LIST=lapply(1:4, function(x) data.frame(rbind(total.ratio4[[x]][-(1)], total.ratio2[[x]][-(1)]), row.names=row))
The following code gives me red and green font colors based on the value on the cell, and it works like a charm:
formattable(TABLE_LIST[[1]], list(area(,-(c(5,10)))~formatter("span", style=x~style(color=ifelse(x>1,"red","green"))),area(,(c(5,10)))~formatter("span", style=x~style(color=ifelse(x>1,"green","red")))))
However, I need COLOR AND comma separated numbers. My failed attempt is:
formattable(TABLE_LIST[[1]], list(area(,-(c(5,10)))~formatter("span", style=x~style(color=ifelse(x>1,"red","green"))),area(,(c(5,10)))~formatter("span", style=x~style(color=ifelse(x>1,"green","red"),digits(x,2))),
area(1:2,1:10)~formatter("span",x~ style(digits(x,2)))))
This code works well, but erases the formatting of the color. I do not know what else to do.
I have to mention I cannot change the original data.frame without messing everything up. So I gotta make the changes on table_list or formattable. Thank you.
I think I solved it. So I will share this small knowledge to people who may have the same problems as me:
formattable(TABLE_LIST[[1]],
list(
area(,-(c(5,10)))~formatter("span",
style=x~style(color=ifelse(x>1,"red","green")),
x~style(digits(x,4))),
area(,(c(5,10)))~formatter("span",
style=x~style(color=ifelse(x>1,"green","red")),
x~style(digits(x,4)))))
Basically, inside the same formatter, on the level of style, add a comma and x~style.
I try to implement a nested tab module the following way.
By clicking on a .tabs__menu item I want to get the next .tabs__contents to display the correct entry.
I've prepared a codepen with markup and leave out all unimportant code so don't be irritated that it's not working. I don't understand why the variable debug2 is 0 and debug3 is 1. I expect debug2 to be 1 as well since I expect the following expression should find the element. Can anyone help me with this?:
.find(".tabs__contents").not(".tabs__contents .tabs__contents");
https://codepen.io/anon/pen/JNLWQp
Thanks in advance and best wishes,
duc
ok I have an assumption why it's not working. It seems that the .not method doesn't starts to search relatively from the given collection but globally. With this statement
.not(".tabs__contents .tabs__contents")
debug2 finds itself and exclude it from the collection thats why the length is 0.
We are using the following code (generated by php but finally this is on client side)
c3.generate({'bindto':'#b65d3422__salestaffcommunication_xepan_base_view_chart_chart','data':{'keys':{'x':'name','value':['Email','Call','Meeting']},'groups':[['Email','Call','Meeting']],'json':[],'type':'bar'},'axis':{'x':{'type':'category'},'rotated':true},'onrendered':function(ev,ui){$(".widget-grid").masonry({'itemSelector':'.widget'})}});
c3.generate({'bindto':'#f67e14d8__t_masscommunication_xepan_base_view_chart_chart','data':{'keys':{'x':'name','value':['Newsletter','TeleMarketing']},'groups':[['Newsletter','TeleMarketing']],'json':[],'type':'bar'},'axis':{'x':{'type':'category'},'rotated':true},'onrendered':function(ev,ui){$(".widget-grid").masonry({'itemSelector':'.widget'})}});
c3.generate({'bindto':'#517df254__ableworkforce_xepan_base_view_chart_chart','data':{'columns':[['present',11.111111111111]],'type':'gauge'},'color':{'pattern':['#FF0000','#F97600','#F6C600','#60B044'],'threshold':{'values':[30,60,90,100]}},'onrendered':function(ev,ui){$(".widget-grid").masonry({'itemSelector':'.widget'})}});
And last graph is not drawn. showing
SyntaxError (DOM Exception 12): The string did not match the expected pattern.
However, I can run ANY two and it works fine. that means all code is perfect but once second one is drawn ( no matter in which order). Third one doesn't draws.
Is it any known bug, or any workaround known.
Using v0.4.11 of c3 from c3js.org
Here is my jsfiddle
https://jsfiddle.net/2yy2mjaf/1/
Thank you.
IDs cannot start with a number, which is the case of your third ID.
The simple solution is just adding a letter to it:
'bindto':'#a517df254_ //just put an "a" before the number here
Here is your fiddle: https://jsfiddle.net/7dkLdg32/
The children functions returns the set of the member.
But I need the children of several members.
The problem is, that I can't use Union to make it work like that:
Union([Geography].[Geography].[USA].children,[Geography].[Geography].[Canada].children)
I don't know how many member it will be... So I actually would need all children of a set of members.
like:
([Geography].[Geography].[USA],[Geography].[Geography].[Canada],[Geography].[Geography].[GB]).children
Is there a function like that?
I couldn't answer my question and so I just edit it. With the help of DHN's answer and some brain work I found a solution I could use:
Except(DRILLDOWNLEVEL( {[Geography].[Geography].[USA],[Geography].[Geography].[Canada]},,0 ),
{[Geography].[Geography].[USA],[Geography].[Geography].[Canada]})
That does work for me.
Explanation: I drilldown the elements the tool provides me, which returns children plus parents and then I use DHN's idea and except the parents so clean the list up a bit.
Hopefully it is understandable.
You can use the Descendants method (the fourth form of the description linked uses a set as its first argument. Thus,
Descendants( {
[Geography].[Geography].[USA],
[Geography].[Geography].[Canada],
[Geography].[Geography].[GB]
},
1,
SELF
)
should deliver exactly what you want.
Well actually, you could use a Crossjoin to get the set you want.
Something like
[Geography].[Geography].[USA] * [Geography].[Geography].[Canada] * [Geography].[Geography].[GB]
But this is only a proper solution, if you have only a few different search criteria.
Alternatively, you could use Except to remove those criteria you're not interested in. E.g.
Except([Geography].[Geography].children, [Geography].[Geography].[Germany])
This would give you the whole content of the [Geography] dimension, except the one of [Germany].
Hope this helps a bit.
Edit after comment of TO
Ok, this wasn't part of your question, but I think what you need is the MemberToStr() function. Please find the doc here.
I think something like this should do the trick.
with member [Measures].[Cities]
as membertostr([Geography].[Geography].members.children)
select [Measures].[Cities] on 0
from [WhatEverYourCubeNameIs]
where (
[Geography].[Geography].[USA],
[Geography].[Geography].[Canada]
)
Please note that this query is totally untested. I also may have lost some of my skills, because it's been a while, since I used mdx. You will also have to create the query dynamically, since the selection seems to be user dependant. But I'm sure that you're aware of it. ;)