If I have a list
a=[1,0,0,1,0,1,1,1,0,1,0,0]
I want to find the index of 0 and 1 respectively, say in this case,
index_0 = [1,2,4,8,10,11]
index_1 = [0,3,5,6,7,9]
is there an efficient way to do this?
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possible duplicate of How to find all occurrences of an element in a list?simopopov– simopopov2014-12-07 16:46:22 +00:00Commented Dec 7, 2014 at 16:46
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3 Answers
index_0 = [i for i, v in enumerate(a) if v == 0]
index_1 = [i for i, v in enumerate(a) if v == 1]
Or with numpy:
import numpy as np
a = np.array(a)
index_0 = np.where(a == 0)[0]
index_1 = np.where(a == 1)[0]
using itertools.compress:
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_1 = [x for x in itertools.compress(range(len(a)),a)]
>>> index_1
[0, 3, 5, 6, 7, 9]
>>> index_0 = [x for x in itertools.compress(range(len(a)),map(lambda x:not x,a))]
>>> index_0
[1, 2, 4, 8, 10, 11]
you can achieve using one for loop: for better and more efficiency
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_0 = []
>>> index_1 = []
>>> for i,x in enumerate(a):
... if x: index_1.append(i)
... else: index_0.append(i)
...
>>> index_0
[1, 2, 4, 8, 10, 11]
>>> index_1
[0, 3, 5, 6, 7, 9]
Comments
Another way to do it is:
import os
a = [1,0,0,1,0,1,1,1,0,1,0,0]
index_0 = []
index_1 = []
aux = 0
for i in a:
if i == 0:
index_0.append(aux)
aux += 1
else:
index_1.append(aux)
aux += 1
print index_0
print index_1
