Pandas - Iterate Over Rows And Calculate - Faster

I already have a solution -but it is very slow (13 minutes for 800 rows). here is an example of the dataframe: import pandas as pd d = {'col1': [20,23,40,41,48,49,50,50], 'col2': [

Solution 1:

IIUC, you can do:

df['overlap_count'] = 0
for i in range(1,start_at_nr+1):
    df['overlap_count'] += df['col1'].le(df['col2'].shift(i))

# mask the first few rows
df.iloc[:start_at_nr, -1] = np.nan

Output:

   col1  col2  overlap_count
02039NaN12332NaN24042NaN341501.0448631.0549682.0650683.0750693.0

Takes about 11ms on for 800 rows and start_at_nr=3.

Solution 2:

You basically compare the current value of col1 to previous 3 rows of col2 and starting the compare from row 3. You may use shift as follow

n = 3
s = ((pd.concat([df.col2.shift(x) for x inrange(1,n+1)], axis=1) >= df.col1.values[:,None])
        .sum(1)[3:])

or

s = (pd.concat([df.col2.shift(x) for x in range(1,n+1)], axis=1).ge(df.col1,axis=0)
                                                                .sum(1)[3:])


Out[65]:
3    1
4    1
5    2
6    3
7    3
dtype: int64

To get your desired output, assign it back to df and fillna

n = 3
s = (pd.concat([df.col2.shift(x) for x in range(1,n+1)], axis=1).ge(df.col1,axis=0)
                                                                .sum(1)[3:])
df_final = df.assign(overlap_count=s).fillna('x')

Out[68]:
   col1  col2 overlap_count
0    20    39             x
1    23    32             x
2    40    42             x
3    41    50             1
4    48    63             1
5    49    68             2
6    50    68             3
7    50    69             3

Solution 3:

You could do it with .apply() in a single statement as follows. I have used a convenience function process_row(), which is also included below.

df.assign(OVERLAP_COUNT = (df.reset_index(drop=False).rename(
                                columns={'index': 'ID'})).apply(
                                    lambda x: process_row(x, df, offset=3), axis=1))

For More Speed: In case you need more speed and are processing a lot of rows, you may consider using swifter library. All you have to do is:

• install swifter: pip install swifter.
• import the library as import swifter.
• replace any .apply() with .swifter.apply() in the code-block above.

Solution in Detail

#!pip install -U swifter#import swifter import numpy as np
import pandas as pd

d = {'col1': [20,23,40,41,48,49,50,50], 'col2': [39,32,42,50,63,68,68,69]}
df = pd.DataFrame(data=d)

defprocess_row(x, df, offset=3):
    value = (df.loc[x.ID - offset:x.ID - 1, 'col2'] >= df.loc[x.ID, 'col1']).sum() if (x.ID >= offset) else'x'return value

# Use df.swifter.apply() for faster processing, instead of df.apply()
df.assign(OVERLAP_COUNT = (df.reset_index(drop=False, inplace=False).rename(
                                columns={'index': 'ID'}, inplace=False)).apply(
                                    lambda x: process_row(x, df, offset=3), axis=1))

Output:

   col1  col2 OVERLAP_COUNT
0    20    39             x
1    23    32             x
2    40    42             x
3    41    50             1
4    48    63             1
5    49    68             2
6    50    68             3
7    50    69             3