根据项条件在两个列表中查找索引

Question

Lets say I have two lists. They are lists of ratings of books on a scale from -5, to 5.

I want to know when list1's element is >= 1 and list2's element == 0 , so for example.

list1 = [3, 3, 1, 0, 3, 0, 3, 0, 0, -3, 0, 5, 3, 0, 1, 0, 0, 5, 3, 0, 0, 0, 0, 1, 0, 3, 0, 1, 0, 0, 3, 5, 3, 3, 0, 0, 0, 5, 0, 5, 0, 3, 3, 0, -3, 0, 0, 5, 1, 5, 3, 0, 3, 0, 0]
list2 = [5, 0, 0, 0, 0, 0, 5, 0, 0, 1, 0, 5, 3, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 5, 0, 0, 0, 0, 5, 5, 5, 3, 0, 0, 0, 3, 0, 0, 0, 5, 3, 0, 0, 0, 0, 5, 0, 5, 3, 0, 0, 0, 0]

list1[1] = 3 and list2[1] = 0 , I want to be able to find all the different indexes of where this happens at.

Sorry if this is confusing but I don't really know how else to word this.

Answer 1

>>> [i for i, v in enumerate(list1) if v>=1 and list2[i]==0]
[1, 2, 4, 14, 18, 27, 39, 48, 52]

Answer 2

Another variant:

>>> [i for i, (l1, l2) in enumerate(zip(list1, list2)) if l1 >= 1 and l2 == 0]
[1, 2, 4, 14, 18, 27, 39, 48, 52]

Answer 3

>>>idx_list = [i for i in range(len(list1)) if list1[i] > 1 and list2[i] == 0]

Answer 4

I found this more readable.

>>> from itertools import count
>>> [i for i,one,two in zip(count(0), list1, list2) if one >= 1 and two == 0]
[1, 2, 4, 14, 18, 27, 39, 48, 52]

And here's the itertools.count doc .

Answer 5

Using NumPy arrays, this is do-able with logical indexing:

 import numpy as np
 list1 = np.array([1, -1, 0, 0, 1])
 list2 = np.array([0, 5, 0, 0, 0])

 # Option 1, multiply the logicals together.
 inds = np.where( (list1 >= 1)*(list2 == 0) )[0]

 # Option 2, pure logicals.
 inds = np.where( (list1 >= 1) & (list2 == 0) )[0]

Now inds[0] = 0 and inds[1] = 4 .