Check the presence of a number after nth digit and then filter your dataframe in Python

Question

My data consists of a date column, which looks like –

Date
20221201
20221202
20221203
20221101
20221102

I need to filter the dataframe only for the month of december, I tried something like –

df = df[df['Date'][4:6] == 12]

its not working

Asked By: Zeren

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Source

Answer 1

I would convert strings into actual dates, then extract month and check if it is equal to desired following way

import pandas as pd
df = pd.DataFrame({"Date":["20221201","20221202","20221203","20221101","20221102"]})
df["Date_object"] = pd.to_datetime(df["Date"],format="%Y%m%d")
df2 = df[df["Date_object"].dt.month==12]
print(df2)

gives output

       Date Date_object
0  20221201  2022-12-01
1  20221202  2022-12-02
2  20221203  2022-12-03

Answered By: Daweo

Answer 2

See @Daweo’s answer for the clean way to do this.

If you actually want to ‘slice’ an integer (not recommended with dates), you can use a floor division (//) in combination with the modulus (%).

df = df[(df['Date'] // 100) % 100 == 12]

Explanation:
// 100 ‘removes’ the last two digits while % 100 only keeps the last two digits.

20221201 // 100 = 202212
202212 % 100 = 12

Answered By: Erik Fubel

Check the presence of a number after nth digit and then filter your dataframe in Python

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