CODEWITHSUNDEEP
python
Srivatsan
Srivatsan

Reputation: 9363

Appending a value to an empty list according to conditions

I have a table with many columns. I am only interested in two columns. Lets call them A and B. I would like to create an empty column, say C = [ ] and append values to it according to some conditions. 

A   B   
1   10
0   11
0   12
1   09
0   14
0   13
1   12

Lets take the above table as an example. If the value in column A==1 then append the value of B for which A==1 to column C (the empty list) until the next A==1 appears in column A. For the next value of 1 in column A, do the above.

So I would want my column C to look like this: 

A   B   C 
1   10  10
0   11  10
0   12  10
1   09  09
0   14  09
0   13  09
1   12  12

So as you can see, for the first A==1, C has the value of B for which A==1 until the next A==1 appears.

I was trying something like this, but it is not working:

for i in A:
    if i == 1:
        C.append(B[A==i])

But this loops through my entire column A and appends many lists.

How can I go about doing it?

Upvotes: 0

Views: 595

Answers (1)

Cory Kramer
Cory Kramer

Reputation: 117856

A = [1,0,0,1,0,0,1]
B = [10,11,12,9,14,13,12]

C = []
temp = B[0]   # Initialize a temporary variable to the first item in B

for i in range(len(B)):
    if A[i] == 1:
        temp = B[i]  # Only update temp if 1 was in the A column
    C.append(temp)

>>> C
[10, 10, 10, 9, 9, 9, 12]

Walking through a few iterations of the for loop:

i = 0, A[0] is 1, temp becomes B[0] which is 10
i = 1, A[1] is 0, temp stays B[0] which is still 10
i = 2, A[2] is 0, temp stays B[0] which is still 10
i = 3, A[3] is 1, temp becomes B[3] which is 9
... etc

Upvotes: 2

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