i got dataframe with column like this:
Date
3 mins
2 hours
9-Feb
13-Feb
the type of the dates is string for every row. What is the easiest way to get that dates into integer unixtime ?
One idea is convert columns to datetimes and to timedeltas:
df['dates'] = pd.to_datetime(df['Date']+'-2020', format='%d-%b-%Y', errors='coerce')
times = df['Date'].replace({'(\d+)\s+mins': '00:\\1:00',
'\s+hours': ':00:00'}, regex=True)
df['times'] = pd.to_timedelta(times, errors='coerce')
#remove rows if missing values in dates and times
df = df[df['Date'].notna() | df['times'].notna()]
df['all'] = df['dates'].dropna().astype(np.int64).append(df['times'].dropna().astype(np.int64))
print (df)
Date dates times all
0 3 mins NaT 00:03:00 180000000000
1 2 hours NaT 02:00:00 7200000000000
2 9-Feb 2020-02-09 NaT 1581206400000000000
3 13-Feb 2020-02-13 NaT 1581552000000000000
Related
I have a Date column in my dataframe having dates with 2 different types (YYYY-DD-MM 00:00:00 and YYYY-DD-MM) :
Date
0 2023-01-10 00:00:00
1 2024-27-06
2 2022-07-04 00:00:00
3 NaN
4 2020-30-06
(you can use pd.read_clipboard(sep='\s\s+') after copying the previous dataframe to get it in your notebook)
I would like to have only a YYYY-MM-DD type. Consequently, I would like to have :
Date
0 2023-10-01
1 2024-06-27
2 2022-04-07
3 NaN
4 2020-06-30
How please could I do ?
Use Series.str.replace with to_datetime and format parameter:
df['Date'] = pd.to_datetime(df['Date'].str.replace(' 00:00:00',''), format='%Y-%d-%m')
print (df)
Date
0 2023-10-01
1 2024-06-27
2 2022-04-07
3 NaT
4 2020-06-30
Another idea with match both formats:
d1 = pd.to_datetime(df['Date'], format='%Y-%d-%m', errors='coerce')
d2 = pd.to_datetime(df['Date'], format='%Y-%d-%m 00:00:00', errors='coerce')
df['Date'] = d1.fillna(d2)
I have a dataframe with two columns (date and days).
df = pd.DataFrame({'date':[2020-01-31, 2020-01-21, 2020-01-11], 'days':[1, 2, 3]})
I want to have a third column (date_2) for which to substract the number of days from the date. Therefore, date_2 would be [2020-01-30, 2020-01-19, 2020-01-8].
I know timedelta(days = i) but I cannot give it the content of df['days'] as i in pandas.
Use to_timedelta with unit=d and subtract
>>pd.to_datetime(df['date'])-pd.to_timedelta(df['days'],unit='d')
0 2020-01-30
1 2020-01-19
2 2020-01-08
dtype: datetime64[ns]
Use to_datetime for datetimes and subtract by Series.sub with timedeltas created by to_timedelta:
df['new'] = pd.to_datetime(df['date']).sub(pd.to_timedelta(df['days'], unit='d'))
print (df)
date days new
0 2020-01-31 1 2020-01-30
1 2020-01-21 2 2020-01-19
2 2020-01-11 3 2020-01-08
I have a dataframe (df) with a column 'Date of birth' column:
Date of birth
0 1957-04-30 00:00:00
1 1966-11-10 00:00:00
2 1966-11-10 00:00:00
3 1962-03-28 00:00:00
4 1958-10-28 00:00:00
5 1958-06-04 00:00:00
How can I reformat the column to a date only format? After I reformat I'm going to work out age from a specific date:
Date of birth
0 1957-04-30
1 1966-11-10
2 1966-11-10
3 1962-03-28
4 1958-10-28
5 1958-06-04
I have tried using
df["Date of birth"] = pd.to_datetime(df['Date of birth'], format='%d%b%Y')
df["Date of birth"] = df["Date of birth"].dt.strftime('%m/%d/%Y')
but with no joy.
After the column becomes a date, use date accessor to access it.
df["Date of birth"] = pd.to_datetime(df['Date of birth']).dt.date
I have a dataframe column which looks like this :
It reads M:S.MS. How can I convert it into a M:S:MS timeformat so I can plot it as a time series graph?
If I plot it as it is, python throws an Invalid literal for float() error.
Note
: This dataframe contains one hour worth of data. Values between
0:0.0 - 59:59.9
df = pd.DataFrame({'date':['00:02.0','00:05:0','00:08.1']})
print (df)
date
0 00:02.0
1 00:05:0
2 00:08.1
It is possible convert to datetime:
df['date'] = pd.to_datetime(df['date'], format='%M:%S.%f')
print (df)
date
0 1900-01-01 00:00:02.000
1 1900-01-01 00:00:05.000
2 1900-01-01 00:00:08.100
Or to timedeltas:
df['date'] = pd.to_timedelta(df['date'].radd('00:'))
print (df)
date
0 00:00:02
1 00:00:05
2 00:00:08.100000
EDIT:
For custom date use:
date = '2015-01-04'
td = pd.to_datetime(date) - pd.to_datetime('1900-01-01')
df['date'] = pd.to_datetime(df['date'], format='%M:%S.%f') + td
print (df)
date
0 2015-01-04 00:00:02.000
1 2015-01-04 00:00:05.000
2 2015-01-04 00:00:08.100
I'm working on a pandas dataframe, one of my column is a date (YYYYMMDD), another one is an hour (HH:MM), I would like to concatenate the two column as one timestamp or datetime64 column, to later use that column as an index (for a time series). Here is the situation :
Do you have any ideas? The classic pandas.to_datetime() seems to work only if the columns contain hours only, day only and year only, ... etc...
Setup
df
Out[1735]:
id date hour other
0 1820 20140423 19:00:00 8
1 4814 20140424 08:20:00 22
Solution
import datetime as dt
#convert date and hour to str, concatenate them and then convert them to datetime format.
df['new_date'] = df[['date','hour']].astype(str).apply(lambda x: dt.datetime.strptime(x.date + x.hour, '%Y%m%d%H:%M:%S'), axis=1)
df
Out[1756]:
id date hour other new_date
0 1820 20140423 19:00:00 8 2014-04-23 19:00:00
1 4814 20140424 08:20:00 22 2014-04-24 08:20:00