Count of 1s in a Pandas DataFrame-CodePudding

After converting a column into 1s and Os based on some criteria using list comprehension and .apply():

x = [0 if df["COLUMN_NAME"][i] > 0 else 1 for i in range(len(df))]
df["PASS"] = df.apply(lambda row: x)

I need to figure out what the longest sequence of 1s I have in my column is, but I can't figure out an easy way to do it. Can someone help me out here. Thanks.

CodePudding user response：

You could adapt this to your problem, as you already have the column in list form.

Longest sequence of consecutive duplicates in a python list

CodePudding user response：

Try: (see at the bottom to understand the method)

import pandas as pd
import numpy as np

np.random.seed(2022)
df = pd.DataFrame({'COL': np.random.choice([0, 1], 20)})

ls1 = df.eq(0).cumsum().loc[df['COL'] == 1].value_counts().max()

Output:

>>> ls1
6

>>> df
    COL
0     1
1     0
2     1
3     0
4     1
5     1
6     0
7     1
8     0
9     0
10    0
11    0
12    1  # <- 1
13    1  # <- 2
14    1  # <- 3
15    1  # <- 4
16    1  # <- 5
17    1  # <- 6
18    0
19    0

Partial result to understand the method:

>>> pd.concat([df, df.eq(0).cumsum().loc[df['COL'] == 1]], axis=1)
    COL  COL
0     1  0.0  # Group 0 contains 1 consecutive 1
1     0  NaN
2     1  1.0  # Group 1 contains 1 consecutive 1
3     0  NaN
4     1  2.0  # Group 2 contains 2 consecutive 1
5     1  2.0
6     0  NaN
7     1  3.0  # Group 3 contains 1 consecutive 1
8     0  NaN
9     0  NaN
10    0  NaN
11    0  NaN
12    1  7.0  # Group 7 contains 6 consecutive 1
13    1  7.0
14    1  7.0
15    1  7.0
16    1  7.0
17    1  7.0
18    0  NaN
19    0  NaN