For instance I have a list A:
A = [100, 200, 300, 200, 400, 500, 600, 400, 700, 200, 500, 800]
And I have list B:
B = [100, 200, 200, 500, 600, 200, 500]
I need to identify the index of elements in B with comparison to A
I have tried:
list_index = [A.index(i) for i in B]
It returns:
[0, 1, 1, 5, 6, 1, 5]
But what I need is:
[0, 1, 3, 5, 6, 9, 10]
How can I solve it?
You can iterate through the enumeration of A to keep track of the indices and yield the values where they match:
A = [100,200,300,200,400,500,600,400,700,200,500,800]
B = [100,200,200,500,600,200,500]
def get_indices(A, B):
a_it = enumerate(A)
for n in B:
for i, an in a_it:
if n == an:
yield i
break
list(get_indices(A, B))
# [0, 1, 3, 5, 6, 9, 10]
This avoids using index() multiple times.
You can try something like this. Move over both lists and append the index when they are equal:
A = [100,200,300,200,400,500,600,400,700,200,500,800]
B = [100,200,200,500,600,200,500]
i, j = 0, 0
list_index = []
while j < len(B):
if B[j] == A[i]:
list_index.append(i)
j += 1
i += 1
print(list_index)
Output:
[0, 1, 3, 5, 6, 9, 10]
You can create a list called indices, and get the first index into it. Then iterate rest of the items in B, then take the slice of A from the last index in indices list and get the index of the item, add it to the last index + 1, then append it back to the indices list.
indices = [A.index(B[0])]
for i,v in enumerate(B[1:]):
indices.append(A[indices[-1]+1:].index(v)+indices[-1]+1)
#indices: [0, 1, 3, 5, 6, 9, 10]