Python, appending to a list using conditions

Question

I have a file with two columns, lets say A and B

A B
1 10
0 11
0 12 
0 15
1 90
0 41

I want to create a new column (a list), lets call the empty list C = []

I would like to loop through A, find if A == 1, and if it is I want to append the value of B[A==1] (10 in the first case) to C until the next A == 1 arrives.

So my final result would be:

A B C
1 10 10
0 11 10
0 12 10
0 15 10
1 90 90
0 41 90

I have tried using the for loop, but only to my dismay:

for a in A:
 if a == 1:
 C.append(B[a==1])
 elif a == 0:
 C.append(B[a==1])

• Malik Brahimi: It shouldn't be a is equal to one. That is the wrong condition for the desired result.
• Malik Brahimi: Are you multiplying 10 to every value with 1?
• user3397243: @MalikBrahimi I am not sure how to implement it using the for loop!!
• Malik Brahimi: What are you trying to do? I don't understand this algorithmically.
• user3397243: @MalikBrahimi I have two lists, given by A and B. I want a new list C as mentioned!! I am not sure what you don't understand. It has to loop through A, find if A == 0, if so append the value of B for which A == 0 to C

Answer 1

You could use another variable to keep the value of the last index in A that had a value of 1, and update it when the condition is met:

temp = 0
for index, value in enumerate(A):
 if value == 1:
 C.append(B[index])
 temp = index
 else:
 C.append(B[temp])

enumerate() gives you a list of tuples with index and values from an utterable. For A, it will be [(0, 1), (1, 0), (2, 0), (3, 0), (4, 1), (5, 0)].

P.S: When you try to address a list using a boolean (B[a == 1]) it will return the item in the first place if the condition is false (B[a != 1] => B[False] => B[0]) or the item in the second place if it's true (B[a == 1] => B[True] => B[1]).

• Don Question: But what if A[0] == 0 ? Still use B[0] or a default value? The OP was not clear about that case.
• user3397243: @DonQuestion A[0] is always 1.. the list proceeds that way!
• OzTamir: @DonQuestion I guess you could add a conditional that will only start appending to C on the first occurrence of 1, but as we don't know what is the purpose of this code I went with the simplest approach.

Answer 2

You may also try using groupby.

Though solution I have come up with looks a bit convoluted to me:

>>> from itertools import izip, groupby, count
>>> from operator import itemgetter
>>> def gen_group(L):
 acc = 0
 for item in L:
 acc += item
 yield acc
>>> [number_out for number,length in ((next(items)[1], 1 + sum(1 for _ in items)) for group,items in groupby(izip(gen_group(A), B), itemgetter(0))) for number_out in repeat(number, length)]
[10, 10, 10, 10, 90, 90]

The idea is to prepare groups and then use them to group your input:

>>> list(gen_group(A))
[1, 1, 1, 1, 2, 2]