Converting all numbers in a python list to 1 or 0

Question

I have a list of integers, it looks like this:

[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]

I need for this list, for every 0, it to stay 0 and every higher number to be 1. Thus converting the above list to:

[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]

I tried to use the following code:

for numbers in list:
   if number==0:
      number=0
   if number>1:
      number=1

however this gives me:

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

-1 for claiming the code you posted gives the output you posted when it does no such thing. If you're having problems with some code, show the actual code you're running. — Wooble, Jul 22 '13 at 13:02
I timed the answers: http://stackoverflow.com/a/17788298/1561176 (Mine is faster) — Inbar Rose, Jul 22 '13 at 13:08

score 14 · Accepted Answer · answered Jul 22 '13 at 12:52

14

>>> 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

answered Jul 22 '13 at 12:52

Roman Pekar

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Timing for different answers available [here](http://stackoverflow.com/a/17788298/1561176) – Inbar Rose Jul 23 '13 at 07:00

Inbar Rose · Answer 2 · 2013-07-22T13:21:07.820

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.

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]

If the numbers are not guaranteed to be non-negative, you'll need to use one of the versions that explicitly compare to 0. — chepner, Jul 22 '13 at 13:21
In which case, as timed above, [Roman Pekar's answer](http://stackoverflow.com/a/17788107/1561176) would be best suited. However, in the OP's question the function is written without a condition for what happens to negative numbers, also - it seems to be this is an attempt to "normalize" a list of values to those `with` a value and those `without` a value. A very common operation in mathematics. If this is not the case, I would be surprised. — Inbar Rose, Jul 22 '13 at 13:23

score 8 · Answer 3 · answered Jul 22 '13 at 12:53

8

myList = [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]
newList = [int(i>0) for i in myList]

OR

newList = [int(bool(i)) for i in myList]

answered Jul 22 '13 at 12:53

inspectorG4dget

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I like the second option with `int(bool(i))` because there's only typecasting and no checking (like `i>0` or some kind of `if`/`else`) in any way. – gitaarik Jul 22 '13 at 12:59
@rednaw: yeah, there's no branching in that one, so it promises to be faster (though I haven't tested the `timeit`s) – inspectorG4dget Jul 22 '13 at 13:01
@inspectorG4dget I timed the answers: http://stackoverflow.com/a/17788298/1561176 (Mine is faster) – Inbar Rose Jul 22 '13 at 13:15
@InbarRose: intriguing. I wonder if the ternary operator is somehow uber optimized – inspectorG4dget Jul 22 '13 at 13:17
1

@inspectorG4dget function calls in python aren't free, the less you make - the better. :) – Inbar Rose Jul 22 '13 at 13:18
The `bool` option requires the numbers to be non-negative (which does seem to be implied, though). – chepner Jul 22 '13 at 13:20
@InbarRose: function calls aren't, but this was type casting and `bool`s were a subclass of `int`s in 2.7, so I thought that would be faster – inspectorG4dget Jul 22 '13 at 13:23

score 2 · Answer 4 · edited May 23 '17 at 12:15

2

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

edited May 23 '17 at 12:15

Community

1
1

answered Jul 22 '13 at 12:54

alecxe

462,703
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list comprehension is faster here, I timed the answers: http://stackoverflow.com/a/17788298/1561176 – Inbar Rose Jul 22 '13 at 13:16
@InbarRose, great, thanks for tests - deserves an upvote! – alecxe Jul 22 '13 at 13:19

blablatros · Answer 5 · 2013-07-22T15:23:11.677

1

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]

edited Jul 22 '13 at 15:23

answered Jul 22 '13 at 12:54

blablatros

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Why was this upvoted? I encourage the upvoter to run the code and see what happens. Hint: `number` is a local variable set to each value in the list, not some magical pointer to the object in the list. Changing it won't change the list. – Wooble Jul 22 '13 at 13:03
My bad, I removed the code completely. – blablatros Jul 22 '13 at 13:14

score 1 · Answer 6 · answered Jul 22 '13 at 13:02

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()

score 0 · Answer 7 · answered Jul 22 '13 at 13:18

0

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

answered Jul 22 '13 at 13:18

rnbguy

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Converting all numbers in a python list to 1 or 0

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