pandas replace only part of a column

Question

Here is my input:

import pandas as pd
import numpy as np

list1 = [10,79,6,38,4,557,12,220,46,22,45,22]
list2 = [4,3,23,6,234,47,312,2,426,42,435,23]

df = pd.DataFrame({'A' : list1, 'B' : list2}, columns = ['A', 'B'])
df['C'] = np.where (df['A'] > df['B'].shift(-2), 1, np.nan)
print (df)

that produces this output:

      A    B    C
0    10    4  NaN
1    79    3  1.0
2     6   23  NaN
3    38    6  NaN
4     4  234  NaN
5   557   47  1.0
6    12  312  NaN
7   220    2  1.0
8    46  426  NaN
9    22   42  NaN
10   45  435  NaN
11   22   23  NaN

What I need to do is to change column 'C' to be a set of three 1's in a row, non-overlapping. The desired output is:

      A    B    C
0    10    4  NaN
1    79    3  1.0
2     6   23  1.0
3    38    6  1.0
4     4  234  NaN
5   557   47  1.0
6    12  312  1.0
7   220    2  1.0
8    46  426  NaN
9    22   42  NaN
10   45  435  NaN
11   22   23  NaN

So, rows 2, 3, and 6 change from NaN to 1.0. Row 7 already has a 1.0 and it is ignored. Rows 8 and 9 need to stay NaN because row 7 is the last entry of the previous set.

I don't know if there is a better way to build column 'C' that would do this at creation.

I have tried several versions of fillna and ffill, none of them worked for me.

It seems very convoluted but I tried to isolate the row id's for each 1.0 with this line:

print (df.loc[df['C'] == 1])

Which correctly outputs this:

     A   B    C
1   79   3  1.0
5  557  47  1.0
7  220   2  1.0

Even though I know that information, I don't know how to proceed from there.

Thank you so much for your help in advance, David

Answer 1

EDIT:

Faster version (thanks to b2002):

ii = df[pd.notnull(df.C)].index
dd = np.diff(ii)
jj = [ii[i] for i in range(1,len(ii)) if dd[i-1] > 2]
jj = [ii[0]] + jj

for ci in jj:
    df.C.values[ci:ci+3] = 1.0

---

First get the indices of all your starting points, i.e. all your points that are 1.0 and have two NaN following, by looking at the differences between the points that are not null in the C column (first index is included by default), then iterate over those indices and use loc to change slices of your C column:

ii = df[pd.notnull(df.C)].index
dd = np.diff(ii)
jj = [ii[i] for i in range(1,len(ii)) if dd[i-1] > 2]
jj = [ii[0]] + jj

for ci in jj:
    df.loc[ci:ci+2,'C'] = 1.0

Result:

      A    B    C
0    10    4  NaN
1    79    3  1.0
2     6   23  1.0
3    38    6  1.0
4     4  234  NaN
5   557   47  1.0
6    12  312  1.0
7   220    2  1.0
8    46  426  NaN
9    22   42  NaN
10   45  435  NaN
11   22   23  NaN

Answer 2

list1 = [10,79,6,38,4,557,12,220,46,22,45,22]
list2 = [4,3,23,6,234,47,312,2,426,42,435,23]

df = pd.DataFrame({'A' : list1, 'B' : list2}, columns = ['A', 'B'])
df['C'] = np.where (df['A'] > df['B'].shift(-2), 1, np.nan)

      A    B    C
0    10    4  NaN
1    79    3  1.0
2     6   23  NaN
3    38    6  NaN
4     4  234  NaN
5   557   47  1.0
6    12  312  NaN
7   220    2  1.0
8    46  426  NaN
9    22   42  NaN
10   45  435  NaN
11   22   23  NaN

make an array from sequence:

a = np.array(df.C)

This function will test segments of an array for matching patterns and will replace segments which match with another pattern. Previously matched segments will not be considered for future matches (the filler numbers are greater than one).

def fill_segments(a, test_patterns, fill_patterns):
    # replace nans with zeros so fast numpy array_equal will work
    nan_idx = np.where(np.isnan(a))[0]
    np.put(a, nan_idx, 0.)
    col_index = list(np.arange(a.size))
    # loop forward through sequence comparing segment patterns
    for j in np.arange(len(test_patterns)):
        this_pattern = test_patterns[j]
        snip = len(this_pattern)
        rng = col_index[:-snip + 1]
        for i in rng:
            seg = a[col_index[i: i + snip]]
            if np.array_equal(seg, this_pattern):
                # when a match is found, replace values in array segment
                # with fill pattern
                pattern_indexes = col_index[i: i + snip]
                np.put(a, pattern_indexes, fill_patterns[j])
    # convert all fillers to ones
    np.put(a, np.where(a > 1.)[0], 1.)
    # convert zeros back to nans
    np.put(a, np.where(a == 0.)[0], np.nan)

    return a

Patterns to be replaced:

p1 = [1., 1., 1.]
p2 = [1., 0., 1.]
p3 = [1., 1., 0.]
p4 = [1., 0., 0.]

And corresponding fill patterns:

f1 = [5., 5., 5.]
f2 = [4., 4., 4.]
f3 = [3., 3., 3.]
f4 = [2., 2., 2.]

make test_patterns and fill_patterns inputs

patterns = [p1, p2, p3, p4]
fills = [f1, f2, f3, f4]

run function:

a = fill_segments(a, patterns, fills)

assign a to column C

df.C = a

df:

      A    B    C
0    10    4  NaN
1    79    3  1.0
2     6   23  1.0
3    38    6  1.0
4     4  234  NaN
5   557   47  1.0
6    12  312  1.0
7   220    2  1.0
8    46  426  NaN
9    22   42  NaN
10   45  435  NaN
11   22   23  NaN

The patterns and fills may need to be adjusted/added to depending on the way the input column is initially populated and the specific result sequence rules.