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Integer manipulation with python

I have an integer that looks like

a = 1010101010010100

that I would like to turn into something like this

list = [[1,0,1,0],[1,0,1,0],[1,0,0,1],[0,1,0,0]]

What I was trying to do is something like

b = str(a)
list1 = [];
for x in b:
    if b.index(x)==0:
        tmp_list = [];
    if b.index(x)!=0%3:
        tmp_list.append(x):
    if b.index(x)==0%3 & b.index(x)!=0:
        list1.append(tmp_list)
        tmp_list=[];

I am not getting the result as expected.

+3


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9 replies


If you want to split your sequence in groups of 4, you can simply do:

>>> b='1010101010010100'

>>> [b[i:i+4] for i in range(0, len(b), 4)]
['1010', '1010', '1001', '0100']

or wrap the element in list(..)

if you want exactly a list of lists:



>>> [list(b[i:i+4]) for i in range(0, len(b), 4)]
[['1', '0', '1', '0'], ['1', '0', '1', '0'], ['1', '0', '0', '1'], ['0', '1', '0', '0']]

also, if the items in the sublists are to be int

:

>>> [list(int(c) for c in b[i:i+4]) for i in range(0, len(b), 4)]
[[1, 0, 1, 0], [1, 0, 1, 0], [1, 0, 0, 1], [0, 1, 0, 0]]
+1


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Here is my solution:

a = 1010101010010100
k = []
m = [int(i) for i in str(a)]

for j in range(0, len(m), 4):
    k.append(m[j:j+4])


output:

[[1, 0, 1, 0], [1, 0, 1, 0], [1, 0, 0, 1], [0, 1, 0, 0]]
0


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You have a lot of problems in your code -

  • I think what you want to do is - b.index(x)%3 == 0

    and not b.index(x)!=0%3

    (which doesn't make much sense and is always false, unless b.index (x) is 0.

  • You have a colon in the line - tmp_list.append(x):

    this is a syntax error (you should probably remove it).

  • &

    is a bitwise and operator in python, I suppose you should use and

    instead &

    .

  • list.index(<element>)

    gives the first index at which the element exists, since your string only consists of 1s and 0s, this will never go past index 2. Instead, you should consider using enumerate

    that will give you the index as well as the value of the element.

Sentence -

  • You don't have to have a condition if b.index(x)==0:

    , you can just define temp_list outside of the for loop.

  • Use if/elif

    (instead of all if

    s).

Sample code -

b = str(a)
list1 = []
tmp_list = []
for i,x in enumerate(b):
    if i%4!=0 or i==0:
        tmp_list.append(x)
    else:
        list1.append(tmp_list)
        tmp_list=[]
        tmp_list.append(x)

output -

[['1', '0', '1', '0'], ['1', '0', '1', '0'], ['1', '0', '0', '1']]
0


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b = str(a)
n = 4
[list([b[i:i+n]) for i in range(0,len(b),n)]
0


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One of the effective solutions could be this: itertools receipes

It returns an iterator, so it should be pretty fast.

from itertools import izip_longest

def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

numbers = str(1010101010010100)
nth = 3
result = grouper(numbers, nth)
0


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import math

a = 1010101010010100
res = []

long = math.ceil(math.log10(a))
digits = [0] * (4 - long % 4) if long % 4 else []
digits.extend([int(i) for i in str(a)])

for i in xrange(0, len(digits), 4):
    res.append(digits[i:i+4])

print(res)
0


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One of them should do (will give different results in case your number length is% 4! = 0):

num = 1010101010010100
[str(num)[i:i+4] for i in range(0, len(str(num)), 4)]
[str(num)[i*4:i*4+4] for i in range(0, len(str(num))//4)]
-1


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List comprehension is your friend (like many others):

a = 1010101010010100
[list(str(a))[i:i+4] for i in range(0,len(str(a)),4)]
-2


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I +1 for a while loop and -1 for an ugly zip:

a = 1010101010010100
result = []

while a:
    a, x = divmod(a, 10)
    result.append(x)

result.reverse()
result = zip(result[0::2], result[1::2], result[2::2], result[3::2])

print(list(result))
-3


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