如何提取特定类别之前的最后3个索引号

Question

UPDATE

I have the following dataset, and I wish to get a list that includes the last three indices before 'YES' label. My dataset:

i            category
0               NO
1               NO
2               NO
3               NO
4               NO
5               YES
6               YES
7               YES
8               NO
9               NO
10              NO
11              YES
12              YES

I expect the outcome to be:

list=[2,3,4,8,9,10]

Please note that YES usually occur in consecutive range of samples (2-6 samples). I wish to get the the last three indices before the first YES in the range.

PS: The dataset was stored in a csv file and I imported by using pandas

Answer 1

Probably not the most pythonic way, but I couldn't think of a way to do this without aa for loop and some slicing, feels like a hacky method:

a = df[((df.category.ne(df.category.shift()))==True) & (df.category == 'YES')].index


indices = []
for x in a:
    indices.append(df.iloc[slice(max(0, x-3), min(x, len(df)))])
new_df = pd.concat(indices) # if you wanted this as a df.

list(new_df.index)


[2, 3, 4, 8, 9, 10]

Answer 2

Let's assume, as you stated on your comment, that there are always at least 3 items before every YES. A possible solution will be

import pandas as pd

flatten = lambda l: [item for sublist in l for item in sublist]

df = pd.DataFrame({"category":['NO', 'NO', 'NO', 'NO', 'NO',
                               'YES', 'NO', 'NO', 'NO', 'NO',
                               'NO','YES','NO']})
# take only indices where YES occurs
idx = df[df["category"]=="YES"].index

# for every i in idx take the previuos 3 indices
lst = [list(range(i-3, i)) for i in idx]

# flatten lst
lst = flatten(lst)

Answer 3

Here's some code that's easy to read and does what you want. it iterates over the indices of the list and pulls out what you need.

the second for loops is to simply flatten the double list from the result list.

li= ['1','2','3','4','YES','6','7','8','9','0','YES']
result = []
for x in range(len(li)):
  if li[x] is 'YES':
    result.append(li[x-3:x])


final= []
for x in result:
  for y in x:
    final.append(y)

final = ['2', '3', '4', '8', '9', '0']