计算一个列的值连续变化n次以上的次数，连同pandas中的变化、分组依据和条件

Question

I have a pandas dataframe:

import pandas as pd
foo = pd.DataFrame({'id': ['a','a','a','a','a','b','b','b','b','b', 'c','c','c','c'], 
                'week': [1,2,3,4,5,3,4,5,6,7,1,2,3,4],
                'col': [1,1,2,2,1,4,3,3,3,4, 6,6,7,7],
                'confidence': ['h','h','h','l','h','h','h','h','h','h', 'h','h','l','l']})

I want to count how many times ( n_changes ) the value of col changes (together with the previous value( from ) and the new value( to )), only if the new value appears more than or equal n consecutive times and there is at least one 'h' in these n consecutive times. I want to do that by id

In case n=3 , the output should look something like this:

 id from to n_changes
 b  4    3  1

because:

• for b , 3 appears after 4 3 times or more, and in these 3 or more consecutive times there is at least one h

In case n=2 , the output should look something like this:

id from to n
a  1    2  1
b  4    3  1

because:

• for a , 2 appears after 1 2 times or more, and in these 2 or more consecutive times there is at least one h
• for b , 3 appears after 4 2 times or more, and in these 2 or more consecutive times there is at least one h
• c does not appear in the output, because even though 7 appears 2 or more consecutive times after 6 , there is not at least one h in these 2 or more consecutive times

Is there a way to achieve this? Any ideas?

UPDATE

I have tried this for n=2

test['next_col'] = test.groupby(['id'])['col'].transform('shift', periods=-1)
test['next_next_col'] = test.groupby(['id'])['col'].transform('shift', periods=-2)
test['next_confidence'] = test.groupby(['id'])['confidence'].transform('shift', periods=-1)
test['next_next_confidence'] = test.groupby(['id'])['confidence'].transform('shift', periods=-2)
test['n_h'] = (test['next_confidence'] == 'h').apply(lambda x: int(x)) + (test['next_next_confidence'] == 'h').apply(lambda x: int(x))
final_test = test[test.eval('next_col == next_next_col and n_h > =1 and col!= next_col')]
final_test['helper'] = 1
final_test['n'] = final_test.groupby(['id','col','next_col'])['helper'].transform('sum')
final_test[['id','col','next_col', 'n']].rename(columns={'col': 'from',
                                                    'next_col': 'to'})

which gives as output

id  from    to  n
1   a   1   2.0 1
5   b   4   3.0 1

which is correct. But is there a more efficient way of doing it?

Answer 1

Here is a way to do this. The key idea is to establish a run_no value that identifies each runs of consecutive col values (within a given id ). Note that there is no groupby(...).apply(some_python_function) , and thus is likely to be quite fast even on large df .

# first, let's establish a "run_no" which is distinct for each
# run of same 'col' for a given 'id'.
# we also set a 'is_h' for later .any() operation, plus a few useful columns:

cols = ['id', 'col']
z = df.assign(
    from_=df.groupby('id')['col'].shift(1, fill_value=-1),
    to=df['col'],
    run_no=(df[cols] != df[cols].shift(1)).any(axis=1).cumsum(),
    is_h=df['confidence'] == 'h')

# next, make a mask that selects the rows we are interested in
gb = z.groupby(['id', 'run_no'])
mask = (gb.size() >= n) & (gb['is_h'].any() & (gb.first()['from_'] != -1))

# finally, we select according to that mask, and add n_changes:
out = gb.first().loc[mask].reset_index()
out = out.assign(n_changes=out.groupby(['id', 'from_', 'to']).size().values)[['id', 'from_', 'to', 'n_changes']]

Outcome, with n = 2 :

>>> out
  id  from_  to  n_changes
0  a      1   2          1
1  b      4   3          1

And with n = 1 :

>>> out
  id  from_  to  n_changes
0  a      1   2          1
1  a      2   1          1
2  b      4   3          1
3  b      3   4          1

Note: if you are interested in the intermediary values, you may of course inspect z (which is independent of n ) and mask (which is dependent on n ). For example, for z :

>>> z
   id  week  col confidence  from_  to  run_no   is_h
0   a     1    1          h     -1   1       1   True
1   a     2    1          h      1   1       1   True
2   a     3    2          h      1   2       2   True
3   a     4    2          l      2   2       2  False
4   a     5    1          h      2   1       3   True
5   b     3    4          h     -1   4       4   True
6   b     4    3          h      4   3       5   True
7   b     5    3          h      3   3       5   True
8   b     6    3          h      3   3       5   True
9   b     7    4          h      3   4       6   True
10  c     1    6          h     -1   6       7   True
11  c     2    6          h      6   6       7   True
12  c     3    7          l      6   7       8  False
13  c     4    7          l      7   7       8  False