CODEWITHSUNDEEP
pythonpandas
sadAndClueless72
sadAndClueless72

Reputation: 69

Find date name in datetime column with user input as weekday name as a 'String'

I have a datetime column called 'Start Time'

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I am trying to find entries with a specific weekday name based on user input.

The user input is a String and can be Sunday, Monday,...Saturday. So if Sunday is input I want to find all entries that have Sunday as the day, regardless of the month or year.

Here is my code:

user_day = input('Input the name of the day.user_day.')

print(df[df['Start Time'].dt.weekday == dt.datetime.strptime(user_day, '%A')])

The output is: Empty DataFrame Columns: [Unnamed: 0, Start Time, End Time, Trip Duration, Start Station, End Station, User Type] Index: []

Upvotes: 1

Views: 249

Answers (2)

floatingpurr
floatingpurr

Reputation: 8559

  • pandas.Series.dt.weekday: The day of the week with Monday=0, Sunday=6
  • pandas.Series.dt.weekday_name: The name of day in a week (e.g.,: Friday)

Pay attention to user input. You are loading a string, you don't need to convert such a string to datetime using strptime.

You just need to match the weekday_name of the datetimelike in your DF with user input. It's a string matching.

print(df[df['Start Time'].dt.weekday_name == user_day])

Upvotes: 0

jezrael
jezrael

Reputation: 862851

Use weekday_name or strftime with same format:

print(df[df['Start Time'].dt.weekday_name == user_day])

Or:

print(df[df['Start Time'].dt.strftime('%A') == user_day])

Verify:

df = pd.DataFrame({'Start Time':pd.date_range('2015-01-01 15:02:45', periods=10)})
print (df)
           Start Time
0 2015-01-01 15:02:45
1 2015-01-02 15:02:45
2 2015-01-03 15:02:45
3 2015-01-04 15:02:45
4 2015-01-05 15:02:45
5 2015-01-06 15:02:45
6 2015-01-07 15:02:45
7 2015-01-08 15:02:45
8 2015-01-09 15:02:45
9 2015-01-10 15:02:45

user_day = 'Monday'
print(df[df['Start Time'].dt.weekday_name == user_day])
           Start Time
4 2015-01-05 15:02:45

print (df['Start Time'].dt.weekday_name)
0     Thursday
1       Friday
2     Saturday
3       Sunday
4       Monday
5      Tuesday
6    Wednesday
7     Thursday
8       Friday
9     Saturday
Name: Start Time, dtype: object

print (df['Start Time'].dt.strftime('%A'))
0     Thursday
1       Friday
2     Saturday
3       Sunday
4       Monday
5      Tuesday
6    Wednesday
7     Thursday
8       Friday
9     Saturday
Name: Start Time, dtype: object

Upvotes: 2

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