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How can I correctly convert binary characters in list to decimal

I want to create a function which will convert the binary numbers in list array to the decimal numbers. For that I have reversed a list and used for loop to iterate the list items. However I am unable to get correct result. Can anyone help me where am I committing the mistake?
def binary_array_to_number(arr):
#arr = arr.reverse()
arr = arr [::-1]
new_num = 0
for item in arr:
for i in range(len(arr)):
new_num = new_num+item*(2**i)
print(new_num)
binary_array_to_number(arr= [0, 0, 0, 1])

Python has built-in conversion from binary to base-10 numbers using the int() constructor. By converting the array to a string representing the binary number, you can use int to convert it to decimal:
binary_arr = [0, 0, 0, 1]
binary_string = ''.join(str(digit) for digit in binary_arr) # this line simply converts the array to a string representing the binary number [0, 0, 0, 1] -> '0001'
decimal_number = int(binary_string, 2) # the 2 here represents the base for the number - base 2 means binary.
print(decimal_number) # prints '1'

You're not checking if the current bit is 1 or not, so you'll always generate 2**n - 1. Also, you've got two loops running instead of one, which will also lead to an incorrect result.
def binary_array_to_number(arr):
#arr = arr.reverse()
arr = arr [::-1]
new_num = 0
for i in range(len(arr)):
if arr[i]:
new_num += (2**i)
print(new_num)
binary_array_to_number(arr= [0, 1, 0, 1])

Stuck converting binary to integers in python

hi i have this code by which my output comes in binary.
b=np.array(np.zeros((1000,4)))
for i in range(1000):
n = "{0:04b}".format(int(Y[i]))
digits = [int(x) for x in str(n)]
b[i] = digits
this gives an output like this:
[0, 0, 0, 1]
please ignore the commas.. read it like [0 0 0 1]---binary number.
im trying to change it such that it gives integers like:
1
can anyone help me with this.. how to edit my code to bring out integers rather than binary numbers.?

Python's built-in int() will do this for you:
binary_num = ''.join(arr)
regular_num = int(binary_num, 2) # 2 is the base

Can we add all the elements in the variable of a for loop? [duplicate]

This question already has answers here:
How can I multiply all items in a list together with Python?
(15 answers)
Closed 3 years ago.
I am trying to build a program which calculates the factorial of number num1. The question is: how can I multiply all the digits in i?
Here I have used a for loop for generating numbers backward from 5: 5,4,3,2,1,0
But is there any way I can multiply all the digits of the variable i?
for i in range(0,5,-1):
print(i)
The result was as follows:
5
4
3
2
1
Is there any way I can get the product of the numbers. I expect the output to be 120.

Your for-loop is wrong.
range(start, end, step): to get all numbers from 5 down to 1, you need range(5, 0, -1)
You can accumulate the result in a variable within the loop. Start with 1 since that's the multiplicative identity.
retval = 1
for i in range(5, 0, -1):
retval *= i
print(retval)

Note that you can easily compute the product of items in a list using reduce:
>> from functools import reduce
>> numbers = list(range(5,0,-1))
[5, 4, 3, 2, 1]
>> reduce((lambda x, y: x * y), numbers)
120

How to print a matrix using Python [duplicate]

This question already has answers here:
Displaying python 2d list without commas, brackets, etc. and newline after every row
(6 answers)
Closed 6 years ago.
I am trying to create a matrix and then print it using python with the following expected output:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
The code follows:
matrix=[[]]
matrix = [[0 for x in range(4)] for x in range(4)]
print matrix
But the output comes as:
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Please tell me why I'm getting this type of output and help with the correct one

What you are getting is the correct representation of a matrix. You have created a list that contains 4 lists, each list containing 4 '0's. If you want to print it differently then you have some options, but what you are printing is the representation of the above mentioned data structure.
for row in matrix:
print ' '.join(map(str,row))
this should work for you.

Try this:
matrix = "\n".join([" ".join(["0" for x in range(4)]) for x in range(4)])
print matrix
The thing is that you're trying to print a list object instead of a string. This way you join all of the numbers into 1 large string separated with spaces for columns and \n (line endings) for rows.
Keep in mind this method will not work for 2 digit numbers unless you pad the 1 digit with a leading zero or space. In order for it to work in all cases, I suggest using the Tabulate module.

Maybe something like this:
matrix=[[]]
matrix = [[0 for x in range(4)] for x in range(4)]
for a in matrix:
ln = ""
for i in a:
ln += str(i) + " "
print(ln)

create a matrix of binary representation of numbers in python

for programming the Hamming cod in sage ( a compiler based on python) I need to create a matrix in which each column is a binary representation of a number
say Hamming(3) the matrix should look like this
0 0 0 1 1 1 1
0 1 1 0 0 1 1
1 0 1 0 1 0 1
which is the binary represenation of numbers from 1 to 7.
so what i did till now is to convert any given number to it's binary representation:
i made this little function it take two values n and r
and repserent n over r bits
binrep(n,r)
x=n
L=[]
LL=[]
while (n>0):
a=int(float(n%2))
L.append(a)
n=(n-a)/2
while (len(L)<r):
L.append(0)
#print(L)
LL=L[::-1]
return LL
so now i want to collect all the LL i got and make them in one big matrix like the one above

In addition to the other ones, an answer using numpy:
import numpy as np
# we're creating the binary representation for all numbers from 0 to N-1
N = 8
# for that, we need a 1xN matrix of all the numbers
a = np.arange(N, dtype=int)[np.newaxis,:]
# we also need a log2(N)x1 matrix, for the powers of 2 on the numbers.
# floor(log(N)) is the largest component that can make up any number up to N
l = int(np.log2(N))
b = np.arange(l, dtype=int)[::-1,np.newaxis]
# This step is a bit complicated, so I'll explain it below.
print np.array(a & 2**b > 0, dtype=int)
This prints:
[[0 0 0 0 1 1 1 1]
[0 0 1 1 0 0 1 1]
[0 1 0 1 0 1 0 1]]
The line
print np.array(a & 2**b > 0, dtype=int)
does a few things at once. I'll split it up into simpler steps:
# this takes our matrix b and creates a matrix containing the powers of 2
# up to 2^log2(N) == N
# (if N is a power of 2; otherwise, up to the next one below)
powers = 2**b
# now we calculate the bit-wise and (&) for each combination from a and b.
# because a has one row, and b as one column, numpy will automatically
# broadcast all values, so the resulting array has size log2(N)xN.
u = a & powers
# this is almost what we want, but has 4's in the first row,
# 2's in the second row and 1's in the last one.
# one method of getting 1's everywhere is to divide by powers:
print u / powers
# another way is to check where u > 0, which yields an array of bools,
# which we then convert to numbers by turning it into an array of ints.
print np.array(u > 0, dtype=int)

You just have to convert all those binary representations to lists of integers, collect them in a list of lists, and finally transpose that list of lists to get the desired output.
n = 3
binary = []
for i in range(1, 2**n):
s = binrep(i, n)
binary.append(map(int, s))
matrix = zip(*binary)
Or in one line: matrix = zip(*(map(int, binrep(i, n)) for i in range(1, 2**n)))
Result is
[(0, 0, 0, 1, 1, 1, 1),
(0, 1, 1, 0, 0, 1, 1),
(1, 0, 1, 0, 1, 0, 1)]
Also note that there are other ways to convert numbers to binary, e.g. using str.format:
def binrep(n,r):
return "{0:0{1}b}".format(n, r)

Like zoosuck mentioned, I would suggest the bin() function in replacement to your code. To print them in a sequence, assuming that the format is similar:
L = ['10101010', '10101010']
for i in L:
print ' '.join([j for j in i])
In a similar manner, you can append it to a list or a dictionary.