how to fill a list with 0 using python
PythonListPython Problem Overview
I want to get a fixed length list from another list like:
a = ['a','b','c']
b = [0,0,0,0,0,0,0,0,0,0]
And I want to get a list like this: ['a','b','c',0,0,0,0,0,0,0]. In other words, if len(a) < len(b), i want to fill up list a with values from list b up to length of the list b, somewhat similar to what str.ljust does.
This is my code:
a=['a','b','c']
b = [0 for i in range(5)]
b = [a[i] for i in b if a[i] else i]
print a
But it shows error:
File "c.py", line 7
b = [a[i] for i in b if a[i] else i]
^
SyntaxError: invalid syntax
What can i do?
Python Solutions
Solution 1 - Python
Why not just:
a = a + [0]*(maxLen - len(a))
Solution 2 - Python
Use itertools repeat.
>>> from itertools import repeat
>>> a + list(repeat(0, 6))
['a', 'b', 'c', 0, 0, 0, 0, 0, 0]
Solution 3 - Python
Why not just
c = (a + b)[:len(b)]
Solution 4 - Python
To be more explicit, this solution replaces the first elements of b with all of the elements of a regardless of the values in a or b:
a + b[len(a):]
This will also work with:
>>> a = ['a', ['b'], None, False]
>>> b = [0, 1, 2, 3, 4, 5]
>>> a + b[len(a):]
['a', ['b'], None, False, 4, 5]
If you do not want to the result to be longer than b:
>>> a = ['a', ['b'], None, False]
>>> b = [0, 1, 2]
>>> (a + b[len(a):])[:len(b)]
['a', ['b'], None]
Solution 5 - Python
LIST_LENGTH = 10
a = ['a','b','c']
while len(a) < LIST_LENGTH:
a.append(0)
Solution 6 - Python
a+[0]*(len(b) - len(a))
['a', 'b', 'c', 0, 0, 0, 0, 0, 0, 0]
Solution 7 - Python
If you want to fill with mutable values, for example with dicts:
map(lambda x: {}, [None] * n)
Where n is the number of elements in the array.
>>> map(lambda x: {}, [None] * 14)
[{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}]
>>> l = map(lambda x: {}, [None] * 14)
>>> l[0]
{}
>>> l[0]['bar'] = 'foo'
>>> l
[{'bar': 'foo'}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}]
Populating without the lambda would cause every element to be {'bar': 'foo'}!
Solution 8 - Python
Can't you just do:
a = ['a','b','c']
b = [0,0,0,0,0,0,0,0]
c = a + b #= ['a','b','c',0,0,0,0,0,0,0]
Solution 9 - Python
Just another solution:
a = ['a', 'b', 'c']
maxlen = 10
result = [(0 if i+1 > len(a) else a[i]) for i in range(maxlen)]
I like some of the other solutions better.