I need to get the size of bits used in one Integer variable.
like this:
bit_number = 1
bit_number = bit_number <<< 2
bit_size(bit_number) # must return 3 here
the bit_size/1 function is for 'strings', not for integers but, in the exercise, whe need to get the size of bits of the integer.
I'm doing one exercise of compression of an book (Classic Computer Science Problems in Python, of Daivid Kopec) and I'm trying to do in Elixir for study.
This works:
(iex) import Bitwise
(iex) Integer.digits(1 <<< 1, 2) |> length
2
but I'm sure there are better solutions.
(as #Hauleth mentions, the answer here should be 2, not 3)
You can count how many times you can divide it by two:
defmodule Example do
def bits_required(0), do: 1
def bits_required(int), do: bits_required(int, 1)
defp bits_required(1, acc), do: acc
defp bits_required(int, acc), do: bits_required(div(int, 2), acc + 1)
end
Output:
iex> Example.bits_required(4)
3
Related
I've an iterable list of over 100 elements. I want to do something after every 10th iterable element. I don't want to use a counter variable. I'm looking for some solution which does not includes a counter variable.
Currently I do like this:
count = 0
for i in range(0,len(mylist)):
if count == 10:
count = 0
#do something
print i
count += 1
Is there some way in which I can omit counter variable?
for count, element in enumerate(mylist, 1): # Start counting from 1
if count % 10 == 0:
# do something
Use enumerate. Its built for this
Just to show another option...hopefully I understood your question correctly...slicing will give you exactly the elements of the list that you want without having to to loop through every element or keep any enumerations or counters. See Explain Python's slice notation.
If you want to start on the 1st element and get every 10th element from that point:
# 1st element, 11th element, 21st element, etc. (index 0, index 10, index 20, etc.)
for e in myList[::10]:
<do something>
If you want to start on the 10th element and get every 10th element from that point:
# 10th element, 20th element, 30th element, etc. (index 9, index 19, index 29, etc.)
for e in myList[9::10]:
<do something>
Example of the 2nd option (Python 2):
myList = range(1, 101) # list(range(1, 101)) for Python 3 if you need a list
for e in myList[9::10]:
print e # print(e) for Python 3
Prints:
10
20
30
...etc...
100
for i in range(0,len(mylist)):
if (i+1)%10==0:
do something
print i
A different way to approach the problem is to split the iterable into your chunks before you start processing them.
The grouper recipe does exactly this:
from itertools import izip_longest # needed for grouper
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
You would use it like this:
>>> i = [1,2,3,4,5,6,7,8]
>>> by_twos = list(grouper(i, 2))
>>> by_twos
[(1, 2), (3, 4), (5, 6), (7, 8)]
Now, simply loop over the by_twos list.
You can use range loops to iterate through the length of mylist in multiples of 10 the following way:
for i in range(0,len(mylist), 10):
#do something
I'm just starting to learn how to code in Sagemath, I know it's similar to python but I don't have much experience with that either.
I'm trying to add two binary numbers representing fractions. That is, something like
a = '110'
b = '011'
bin(int(a,2) + int(b,2))
But using values representing fractions, such as '1.1'.
Thanks in advance!
If you want to do this in vanilla Python, parsing the binary fractions by hand isn't too bad (the first part being from this answer);
def binstr_to_float(s):
t = s.split('.')
return int(t[0], 2) + int(t[1], 2) / 2.**len(t[1])
def float_to_binstr(f):
i = 0
while int(f) != f:
f *= 2
i += 1
as_str = str(bin(int(f)))
if i == 0:
return as_str[2:]
return as_str[2:-i] + '.' + as_str[-i:]
float_to_binstr(parse_bin('11.1') + parse_bin('0.111')) # is '100.011'
In python you can use the Binary fractions package. With this package you can convert binary-fraction strings into floats and vice-versa. Then, you can perform operations on them.
Example:
>>> from binary_fractions import Binary
>>> sum = Binary("1.1") + Binary("10.01")
>>> str(sum)
'0b11.11'
>>> float(sum)
3.75
>>>
It has many more helper functions to manipulate binary strings such as: shift, add, fill, to_exponential, invert...
PS: Shameless plug, I'm the author of this package.
I have a bytearray filled with "c-type" reversed order data like sint32_t but also sint24_t. A 24-bit signed value needs to be converted into a integer. Python handles negative values as a value with a minus sign and c uses the signed bit to indicate a negative value.
So I came up changing it to 32-bit first:
raw = bytearray('\x89\x00\x23')
val = (ord(raw[0:1]) | (ord(raw[1:2])<<8) | (ord(raw[2:3])<<16) )
if ( (val & 0x00800000L) > 0):
val |= 0xFF000000L
This works however now I have a 32-bit signed value. I still need it to become a negative value in python. So I came up with:
import ctypes
p_val = ctypes.c_int32(val).value
This will convert it in the correct way.
I would like this to be a bit more efficient and faster. Is there any way to rewrite this in something that is much faster. I need to create 7 values like this per iteration. I read something about "memoryview"?
Anyone?
Various options:
import struct
MAX_INT24 = (1<<23)-1
BIAS_INT24 = (1<<24)
def int24(raw):
val = raw[0] | (raw[1] << 8) | (raw[2] << 16)
# val = struct.unpack('<i',raw + b'\x00')[0] # Another option
return val if val <= MAX_INT24 else val - BIAS_INT24
raws = [b'\xff\xff\x7f',
b'\xff\xff\xff',
b'\xfe\xff\xff']
for raw in raws:
print(int24(raw))
print(int.from_bytes(raw,'little',signed=True)) # Python 3 only
Output:
8388607
8388607
-1
-1
-2
-2
Is there an equivalent function of find(A>9,1) from matlab for numpy/scipy. I know that there is the nonzero function in numpy but what I need is the first index so that I can use the first index in another extracted column.
Ex: A = [ 1 2 3 9 6 4 3 10 ]
find(A>9,1) would return index 4 in matlab
The equivalent of find in numpy is nonzero, but it does not support a second parameter.
But you can do something like this to get the behavior you are looking for.
B = nonzero(A >= 9)[0]
But if all you are looking for is finding the first element that satisfies a condition, you are better off using max.
For example, in matlab, find(A >= 9, 1) would be the same as [~, idx] = max(A >= 9). The equivalent function in numpy would be the following.
idx = (A >= 9).argmax()
matlab's find(X, K) is roughly equivalent to numpy.nonzero(X)[0][:K] in python. #Pavan's argmax method is probably a good option if K == 1, but unless you know apriori that there will be a value in A >= 9, you will probably need to do something like:
idx = (A >= 9).argmax()
if (idx == 0) and (A[0] < 9):
# No value in A is >= 9
...
I'm sure these are all great answers but I wasn't able to make use of them. However, I found another thread that partially answers this:
MATLAB-style find() function in Python
John posted the following code that accounts for the first argument of find, in your case A>9 ---find(A>9,1)-- but not the second argument.
I altered John's code which I believe accounts for the second argument ",1"
def indices(a, func):
return [i for (i, val) in enumerate(a) if func(val)]
a = [1,2,3,9,6,4,3,10]
threshold = indices(a, lambda y: y >= 9)[0]
This returns threshold=3. My understanding is that Python's index starts at 0... so it's the equivalent of matlab saying 4. You can change the value of the index being called by changing the number in the brackets ie [1], [2], etc instead of [0].
John's original code:
def indices(a, func):
return [i for (i, val) in enumerate(a) if func(val)]
a = [1, 2, 3, 1, 2, 3, 1, 2, 3]
inds = indices(a, lambda x: x > 2)
which returns >>> inds [2, 5, 8]
Consider using argwhere in Python to replace MATLAB's find function. For example,
import numpy as np
A = [1, 2, 3, 9, 6, 4, 3, 10]
np.argwhere(np.asarray(A)>=9)[0][0] # Return first index
returns 3.
import numpy
A = numpy.array([1, 2, 3, 9, 6, 4, 3, 10])
index = numpy.where(A >= 9)
You can do this by first convert the list to an ndarray, then using the function numpy.where() to get the desired index.
Many random-number generators return floating numbers between 0 and 1.
What's the best and correct way to get integers between a and b?
Divide the interval [0,1] in B-A+1 bins
Example A=2, B=5
[----+----+----+----]
0 1/4 1/2 3/4 1
Maps to 2 3 4 5
The problem with the formula
Int (Rnd() * (B-A+1)) + A
is that your Rnd() generation interval is closed on both sides, thus the 0 and the 1 are both possible outputs and the formula gives 6 when the Rnd() is exactly 1.
In a real random distribution (not pseudo), the 1 has probability zero. I think it is safe enough to program something like:
r=Rnd()
if r equal 1
MyInt = B
else
MyInt = Int(r * (B-A+1)) + A
endif
Edit
Just a quick test in Mathematica:
Define our function:
f[a_, b_] := If[(r = RandomReal[]) == 1, b, IntegerPart[r (b - a + 1)] + a]
Build a table with 3 10^5 numbers in [1,100]:
table = SortBy[Tally[Table[f[1, 100], {300000}]], First]
Check minimum and maximum:
In[137]:= {Max[First /# table], Min[First /# table]}
Out[137]= {100, 1}
Lets see the distribution:
BarChart[Last /# SortBy[Tally[Table[f[1, 100], {300000}]], First],
ChartStyle -> "DarkRainbow"]
X = (Rand() * (B - A)) + A
Another way to look at it, where r is your random number in the range 0 to 1:
(1-r)a + rb
As for your additional requirement of the result being an integer, maybe (apart from using built in casting) the modulus operator can help you out. Check out this question and the answer:
Expand a random range from 1–5 to 1–7
Well, why not just look at how Python does it itself? Read random.py in your installation's lib directory.
After gutting it to only support the behavior of random.randint() (which is what you want) and removing all error checks for non-integer or out-of-bounds arguments, you get:
import random
def randint(start, stop):
width = stop+1 - start
return start + int(random.random()*width)
Testing:
>>> l = []
>>> for i in range(2000000):
... l.append(randint(3,6))
...
>>> l.count(3)
499593
>>> l.count(4)
499359
>>> l.count(5)
501432
>>> l.count(6)
499616
>>>
Assuming r_a_b is the desired random number between a and b and r_0_1 is a random number between 0 and 1 the following should work just fine:
r_a_b = (r_0_1 * (b-a)) + a