Converting all numbers in a python list to 1 or 0 [closed]

Answer 1

A simple list-comprehension

>>> l = [10, 8, 4, 4, 13, 1, 1, 1, 1, 6, 1, 2, 1, 1, 0, 1, 5, 1, 5, 5, 2, 1, 0, 0, 4]
>>> [1 if i else 0 for i in l]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1]

The reason your code was not working is because you are iterating (looping) through each item of the list, but not modifying the item in the original list, only the returned item.

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Timers (In order of speed):

# my method: [1 if i else 0 for i in l]
>>> timeit.Timer('[1 if i else 0 for i in l]', setup='from __main__ import l').repeat()
[1.4960370266417915, 1.457976119063474, 1.457053021255831]
# Roman Pekar: [1 if x > 0 else 0 for x in l]
>>> timeit.Timer('[1 if x > 0 else 0 for x in l]', setup='from __main__ import l').repeat()
[1.5296303022631434, 1.5041486202146075, 1.51295106957906]
# inspectorG4dget: [int(i>0) for i in myList]
>>> timeit.Timer('[int(i>0) for i in l]', setup='from __main__ import l').repeat()
[5.0810576000558285, 4.865218180917736, 4.7859557786252935]
# Haidro: map(int, map(bool, l))
>>> timeit.Timer('map(int, map(bool, l))', setup='from __main__ import l').repeat()
[5.725813168085608, 5.759308116913473, 5.549817013103507]
# alecxe: map(lambda x: int(x > 0), l)
>>> timeit.Timer('map(lambda x: int(x > 0), l)', setup='from __main__ import l').repeat()
[7.054628605392196, 7.291914272244128, 7.223923128993391]
# inspectorG4dget: [int(bool(i)) for i in l]
>>> timeit.Timer('[int(bool(i)) for i in l]', setup='from __main__ import l').repeat()
[8.60473766374804, 8.537255398342722, 8.545150893104449]

Answer 2

Use map:

L = [10, 8, 4, 4, 13, 1, 1, 1, 1, 6, 1, 2, 1, 1, 0, 1, 5, 1, 5, 5, 2, 1, 0, 0, 4]
print map(int, map(bool, L))

Booleans are subclasses of integers, thus False == 0 and True == 1. Calling int() on a bool will return either one or zero for true or false, respectively.

Note that in Python 3, map returns an iterator. So you should call list() on the outside map()

Answer 3

You are using numbers variable when creating a for loop, but then you are comparing number (no 's') variable.

Also, avoid using list as a list name; give it some other name (e.g. a_list, or something descriptive).

Others have already provided you with good answers. Just in case you have a hard time understanding list comprehensions, this is what you tried to do:

old_list = [1, 5, 0, 0, 2]
new_list = []
for each_element in old_list:
    if each_element == 0:
        new_list.append(0)
    else:
        new_list.append(1) # what about negative numbers?

This is equivalent to:

new_list = [0 if each_element == 0 else 1 for each_element in a_list]

You can also "modify" (overwrite) existing list on the fly:

a_list = [0 if each_element == 0 else 1 for each_element in a_list]

Answer 4

How about this:

nums = [10, 8, 4, 4, 13, 1, 1, 1, 1, 6, 1, 2, 1, 1, 0, 1, 5, 1, 5, 5, 2, 1, 0, 0, 4]
for i in range(len(nums)):
    if nums[i]: nums[i] = 1

Answer 5

You can use map instead of list comprehension approach:

>>> l = [10, 8, 4, 4, 13, 1, 1, 1, 1, 6, 1, 2, 1, 1, 0, 1, 5, 1, 5, 5, 2, 1, 0, 0, 4]
>>> map(lambda x: int(x > 0), l)
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1]

Also see: Python List Comprehension Vs. Map

Answer 6

>>> l = [10, 8, 4, 4, 13, 1, 1, 1, 1, 6, 1, 2, 1, 1, 0, 1, 5, 1, 5, 5, 2, 1, 0, 0, 4]
>>> [1 if x > 0 else 0 for x in l]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1]

if you want to change existing list

for i, x in enumerate(l):
    if x >= 1: l[i] = 1