Set the format for pd.to_datetime

Question

Hi already referred to this post but I could not get through my issue. I have a column in my CSV which is string and the sample values are below (note that the month and year positioning are sometimes reversed). What format do I need to set in my to_datetime? I tried all the below approaches

df = pd.read_csv("filename.csv") #Imagine there is a Month column

#[1] df["Month"] = pd.to_datetime(df["Month"])
#[2] df["Month"] = pd.to_datetime(df["Month"], format="%m/%d/%Y")

[Month]
Mar-97
Apr-97
May-97
Jun-97
Nov-00
Dec-00
1-Jan
1-Feb
1-Mar
1-Apr

I get the error

ValueError: day is out of range for month

for [1] and I get

ValueError: time data 'Mar-97' does not match format '%m/%d/%Y' (match)

for [2]. I tried to remove the %d too but no luck. Could you please point me what is going wrong here.

score 1 · Accepted Answer · answered May 20 '18 at 17:49

One way is to use try / except with pd.Series.apply:

s = pd.Series(['Mar-97', 'May-97', 'Nov-00', '1-Jan', '1-Mar'])

def converter(x):
    try:
        return pd.datetime.strptime(x, '%b-%y')
    except ValueError:
        year, month = x.split('-')  # split by delimiter
        x = year.zfill(2) + '-' + month  # %y requires 0-padding
        return pd.datetime.strptime(x, '%y-%b')

res = s.apply(converter)

print(res)

0   1997-03-01
1   1997-05-01
2   2000-11-01
3   2001-01-01
4   2001-03-01
dtype: datetime64[ns]

Since we have defined converter as a function, we can use this directly with pd.read_csv:

df = pd.read_csv('file.csv', parse_dates=['dt_col_name'], date_parser=converter)

Python's strftime directives is a useful reference for constructing datetime format strings.

score 0 · Answer 2 · answered May 20 '18 at 17:57

Not the most elegant, but you might try to fix and order the year and month parts. The below code works:

Recreate your data:

df = pd.DataFrame({"date_str": ['Mar-97', 'Apr-97', 'May-97', 
                                'Jun-97', 'Nov-00', 'Dec-00',
                                '1-Jan', '1-Feb', '1-Mar', '1-Apr']})

Split parts:

df = pd.concat([df, df['date_str'].str.split("-", expand=True)], axis=1)

Organize month and year:

df.loc[df[0].str.len() == 3, 'month'] = df.loc[df[0].str.len() == 3, 0]
df.loc[df[1].str.len() == 3, 'month'] = df.loc[df[1].str.len() == 3, 1]
df.loc[df[0].str.len() != 3, 'year'] = df.loc[df[0].str.len() != 3, 0]
df.loc[df[1].str.len() != 3, 'year'] = df.loc[df[1].str.len() != 3, 1]

Correct years that are only a single digit:

df.loc[df['year'].str.len() == 1, 'year'] = '0' + df.loc[df['year'].str.len() == 1, 'year']

Generate proper date column:

df['date'] = (df['month'] + '-' + df['year']).apply(lambda x: pd.to_datetime(x, format="%b-%y"))

Output:

print(df['date'])

0   1997-03-01
1   1997-04-01
2   1997-05-01
3   1997-06-01
4   2000-11-01
5   2000-12-01
6   2001-01-01
7   2001-02-01
8   2001-03-01
9   2001-04-01
Name: date, dtype: datetime64[ns]

score 0 · Answer 3 · answered Aug 09 '22 at 17:22

Any of this worked for me while using pandas. I fix this problem and other pretty similars of dtypes with this function:

dfReCajatot2["Column dates"] = dfReCajatot2["Column dates""].apply(pd.to_datetime, errors='coerce')

This changes the tipe of each cell instead of all column. Maybe some values throws diferent structure but you can fix that later

Set the format for pd.to_datetime

3 Answers3