Convert the integer to string first, and then use map to apply int on it:
>>> num = 132
>>> map(int, str(num)) #note, This will return a map object in python 3.
[1, 3, 2]
or using a list comprehension:
>>> [int(x) for x in str(num)]
[1, 3, 2]
Answer from Ashwini Chaudhary on Stack Overflowpython - Converting integer to digit list - Stack Overflow
Convert NumPy array to Python list - Stack Overflow
Split up an integer of numbers into arrays of numbers
How to create an integer array in Python? - Stack Overflow
Convert the integer to string first, and then use map to apply int on it:
>>> num = 132
>>> map(int, str(num)) #note, This will return a map object in python 3.
[1, 3, 2]
or using a list comprehension:
>>> [int(x) for x in str(num)]
[1, 3, 2]
There are already great methods already mentioned on this page, however it does seem a little obscure as to which to use. So I have added some mesurements so you can more easily decide for yourself:
A large number has been used (for overhead) 1111111111111122222222222222222333333333333333333333
Using map(int, str(num)):
import timeit
def method():
num = 1111111111111122222222222222222333333333333333333333
return map(int, str(num))
print(timeit.timeit("method()", setup="from __main__ import method", number=10000)
Output: 0.018631496999999997
Using list comprehension:
import timeit
def method():
num = 1111111111111122222222222222222333333333333333333333
return [int(x) for x in str(num)]
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))
Output: 0.28403817900000006
Code taken from this answer
The results show that the first method involving inbuilt methods is much faster than list comprehension.
The "mathematical way":
import timeit
def method():
q = 1111111111111122222222222222222333333333333333333333
ret = []
while q != 0:
q, r = divmod(q, 10) # Divide by 10, see the remainder
ret.insert(0, r) # The remainder is the first to the right digit
return ret
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))
Output: 0.38133582499999996
Code taken from this answer
The list(str(123)) method (does not provide the right output):
import timeit
def method():
return list(str(1111111111111122222222222222222333333333333333333333))
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))
Output: 0.028560138000000013
Code taken from this answer
The answer by Duberly Gonzรกlez Molinari:
import timeit
def method():
n = 1111111111111122222222222222222333333333333333333333
l = []
while n != 0:
l = [n % 10] + l
n = n // 10
return l
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))
Output: 0.37039988200000007
Code taken from this answer
Remarks:
In all cases the map(int, str(num)) is the fastest method (and is therefore probably the best method to use). List comprehension is the second fastest (but the method using map(int, str(num)) is probably the most desirable of the two.
Those that reinvent the wheel are interesting but are probably not so desirable in real use.
Use tolist():
>>> import numpy as np
>>> np.array([[1,2,3],[4,5,6]]).tolist()
[[1, 2, 3], [4, 5, 6]]
Note that this converts the values from whatever numpy type they may have (e.g. np.int32 or np.float32) to the "nearest compatible Python type" (in a list). If you want to preserve the numpy data types, you could call list() on your array instead, and you'll end up with a list of numpy scalars. (Thanks to Mr_and_Mrs_D for pointing that out in a comment.)
c = np.array([[1,2,3],[4,5,6]])
list(c.flatten())
Hello so I want my input to be an integer of 6 numbers โ123456โ but then print them out as 3 arrays with two integer each f.e. โ12โ, โ34โ, โ56โ. Any idea which way is the easiest?
If you are not satisfied with lists (because they can contain anything and take up too much memory) you can use efficient array of integers:
import array
array.array('i')
See here
If you need to initialize it,
a = array.array('i',(0 for i in range(0,10)))
two ways:
x = [0] * 10
x = [0 for i in xrange(10)]
Edit: replaced range by xrange to avoid creating another list.
Also: as many others have noted including Pi and Ben James, this creates a list, not a Python array. While a list is in many cases sufficient and easy enough, for performance critical uses (e.g. when duplicated in thousands of objects) you could look into python arrays. Look up the array module, as explained in the other answers in this thread.