Chris90
Chris90

Reputation: 1998

Creating new column based on time value based if statement

Im trying to create a column that gives the variance or subtraction of two timestamps of two other columns.

def time_delta(df):
    if df['a_time'] > df['b_time']:
        df = (pd.to_datetime(df.a_time) - pd.to_datetime(df.b_time)) / np.timedelta64(1, 'm')
    else:
        df = (pd.to_datetime(df.b_time) - pd.to_datetime(df.a_time)) / np.timedelta64(1, 'm')
    return df

df['C'] = df.apply(time_delta, axis=1)

When I run the apply part of code the cell just keeps running with *, am I missing something?

Thanks so much

Upvotes: 3

Views: 118

Answers (2)

jpp
jpp

Reputation: 164673

Your logic is over-complicated. Row-wise loops, which is what pd.DataFrame.apply represents, should be actively avoided with Pandas. Here, you can convert a timedelta series to seconds, then take the absolute value:

df = pd.DataFrame({'a_time': pd.to_datetime(['2018-01-01 05:32:00', '2018-05-10 20:13:41']),
                   'b_time': pd.to_datetime(['2018-01-01 15:10:05', '2018-05-10 16:09:16'])})

df['C'] = (df['b_time'] - df['a_time']).dt.total_seconds().abs() / 60

print(df)

               a_time              b_time           C
0 2018-01-01 05:32:00 2018-01-01 15:10:05  578.083333
1 2018-05-10 20:13:41 2018-05-10 16:09:16  244.416667

For academic purposes, this is how you would use apply:

def time_delta(row):
    if row['a_time'] > row['b_time']:
        return (row['a_time'] - row['b_time']) / np.timedelta64(1, 'm')
    else:
        return (row['b_time'] - row['a_time']) / np.timedelta64(1, 'm')

df['C'] = df.apply(time_delta, axis=1)

Notice, in both versions, we assume you are starting with datetime series. If this isn't case, make sure you convert to datetime as an initial step:

time_cols = ['a_time', 'b_time']
df[time_cols] = df[time_cols].apply(pd.to_datetime)

Upvotes: 0

Hainan Zhao
Hainan Zhao

Reputation: 2092

Don't assign result to "df", change it to different variable instead.

def time_delta(df):
    if df['a_time'] > df['b_time']:
        res = (pd.to_datetime(df.a_time) - pd.to_datetime(df.b_time)) / np.timedelta64(1, 'm')
    else:
        res = (pd.to_datetime(df.b_time) - pd.to_datetime(df.a_time)) / np.timedelta64(1, 'm')
    return res

Upvotes: 1

Related Questions