I have to remove all specific values from array (if any), so i write:
while value_to_remove in my_array:
my_array.remove(value_to_remove)
Is there more pythonic way to do this, by one command?
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4 Answers
You can try:
filter (lambda a: a != value_to_remove, my_array)
Example:
>>> my_array = ["abc", "def", "xyz", "abc", "pop", "abc"]
>>> filter (lambda a: a != "abc", my_array)
['def', 'xyz', 'pop']
3 Comments
Silas Ray
This seems the best. The only caveat is if he specifically wants to do an in-place edit if the existing list, in which case there's extra allocation being done compared to the loop in his question.
RaSergiy
Thanks. I suppose, that filter is much more faster, then my initial variant:)
georg
filter+lambda is exactly how you are not supposed to do things in python. GvR: "filter(P, S) is almost always written clearer as [x for x in S if P(x)]"clean_array = [element for element in my_array if value_to_remove != element]
my_array = ('a','b','a','c')
value_to_remove = 'a'
>>> clean_array = [element for element in my_array if value_to_remove != element]
>>> clean_array
['b', 'c']
Comments
Use itertools.ifilter:
import itertools
my_array = list(itertools.ifilter(lambda x: x != value_to_remove, my_array))
2 Comments
Silas Ray
What benefit does going through
ifilter give you here? If you just push the result back in to a list anyway, a straight filter seems better.Don't miss that there is an extremely efficient way as follows:
import numpy as Numpy
a = Numpy.array([1,7,3,4,4,6,3,8,7,0,8])
b = Numpy.array(['1','7','3','4','4','6','3','8','7','0','8'])
c = Numpy.array([0.1,9.8,-0.4,0.0,9.8,13.7])
d = Numpy.array(['one','three','five','four','three'])
print a
print a[a!=3]
print b
print b[b!='3']
print c
print c[c!=9.8]
print d
print d[d!='three']
and you get:
>>>
[1 7 3 4 4 6 3 8 7 0 8]
[1 7 4 4 6 8 7 0 8]
['1' '7' '3' '4' '4' '6' '3' '8' '7' '0' '8']
['1' '7' '4' '4' '6' '8' '7' '0' '8']
[ 0.1 9.8 -0.4 0. 9.8 13.7]
[ 0.1 -0.4 0. 13.7]
['one' 'three' 'five' 'four' 'three']
['one' 'five' 'four']
