20160116
Suppose this is the data with datatype integer in a column and now I want to convert it like 2016/01/16 or 2016-01-16 and datatype as date. My column name is system and dataframe is df. How can I do that?
I tried using many date format function but It was not good enough to achieve the answer.
convert using to_datetime, provide the format
then convert to the format of your desire
pd.to_datetime(df['dte'], format='%Y%m%d').dt.strftime('%Y/%m/%d')
0 2016/01/06
Name: dte, dtype: object
Using str.replace we can try:
df["date"] = df["system"].astype(str).str.replace(r'(\d{4})(\d{2})(\d{2})', r'\1/\2/\3', regex=True)
Related
I have a dataframe column in the format of 20180531.
I need to split this properly i.e. I can get 2018/05/31.
This is a dataframe column that I have and I need to deal with it in a datetime format.
Currently this column is identified as int64 type
I'm not sure how efficient it'll be but if you convert it to a string, and the use pd.to_datetime with a .format=..., eg:
df['actual_datetime'] = pd.to_datetime(df['your_column'].astype(str), format='%Y%m%d')
As Emma points out - the astype(str) is redundant here and just:
df['actual_datetime'] = pd.to_datetime(df['your_column'], format='%Y%m%d')
will work fine.
Assuming the integer dates would always be fixed width at 8 digits, you may try:
df['dt'] = df['dt_int'].astype(str).str.replace(r'(\d{4})(\d{2})(\d{2})', r'\1-\2-\3')
I have a csv file with a column "graduated" which either shows the date of graduation, or 0 if there is no graduation yet.
df.dtypes return 'object' for this column, I want to turn all the dates into a '1' (indicating that the person in that column graduated). How can I do that ?
Use pandas.to_datetime to convert dates and convert to boolean series. Then, cast it to int to get the desired result.
pd.to_datetime(df.graduated, errors='coerce').notnull().astype(int)
Basically i am sas developer.
As of now i am doing sas2python migrations.
Before reading to pandas dataframe i have two columns ie,
DATE NAME
01JAN1988 VARUN
11JAN1999 THARUN
After reading to pandas dataframe the DATE columns is automatically read as float values. Now I need to show it as DATE Columns as date9 format
Could you please provide the steps
you can use apply function to convert the values into date objects and datetime module to covert them:
df['DATE'] = df['DATE'].apply(lambda x: datetime.datetime.strptime(x,'%d%b%Y').date())
Output:
DATE NAME
0 1988-01-01 VARUN
1 1999-01-11 THARUN
One of the columns in my pandas dataframe looks like this:
14.3.2019
15.3.2019
16.3.2019
So this is European/German date that I have to convert to USA format:
2019-3-14
2019-3-15
2019-3-16
What is the fastest way to do it, possibly inplace, if I have a large dataset?
Correct answer given by both commenters, posting here faster solution from #QuangHoang.
Casting string column in date type in desired format:
df['date'] = pd.to_datetime(df['date'], format='%d.%m.%Y').dt.strftime('%Y-%m-%d')
I have a dataframe that is called dfactual this dataframe has a column ForeCastEndDate, so
dfactual['ForeCastEndDate'] it contains:
311205
311205
This must be a date in the format 31-12-2005, but the current format is int64. I tried the following:
dfactual['ForeCastEndDate'] = pd.to_datetime(pd.Series(dfactual['ForecastEndDate']))
I tried also to add the format command to it, but it didn't work out the format stays the same, int64.
How should I do it?
You can't use to_datetime with dtypes that are not str so you need to convert the dtype using astype first and then you can use to_datetime and pass the format string:
In [154]:
df = pd.DataFrame({'ForecastEndDate':[311205]})
pd.to_datetime(df['ForecastEndDate'].astype(str), format='%d%m%y')
Out[154]:
0 2005-12-31
Name: ForecastEndDate, dtype: datetime64[ns]