Trying to change multiple columns to the same datatype at once,
columns contain time data like hours minute and seconds, like
And the data
and I'm not able to change multiple columns at once to using pd.to_datetime to only the time format, I don't want the date because, if I do pd.to_datetime the date also gets added to the column which is not required, just want the time
how to convert the column to DateTime and only keep time in the column
First You can't have a datetime with only time in it in pandas/python.
So
Because python time is object in pandas convert all columns to datetimes (but there are also dates):
cols = ['Total Break Time','col1','col2']
df[cols] = df[cols].apply(pd.to_datetime)
Or convert columns to timedeltas, it looks like similar times, but possible working by datetimelike methods in pandas:
df[cols] = df[cols].apply(pd.to_timedelta)
You can pick only time as below:
import time
df['Total Break Time'] = pd.to_datetime(df['Total Break Time'],format= '%H:%M:%S' ).dt.time
Then you can repeat this for all your columns, as I suppose you already are.
The catch is, to convert to datetime and then only picking out what you need.
Related
In the long run, I'm trying to be able to merge different dataframes of data coming from different sources. The dataframes themselves are all a time series. I'm having difficulty with one dataset. The first column is DateTime. The initial data has a temporal resolution of 15 s, but in my code I have it being resampled and averaged for each minute (this is to have the same temporal resolution as my other datasets). What I'm trying to do, is make this 0 key of the datetimes, and then concatenate this horizontally to the initial data. I'm doing this because when I set the index column to 'DateTime', it seems to delete that column (when I export as csv and open this in excel, or print the dataframe, this column is no longer there), and concatenating the 0 (or df1_DateTimes, as in the code below) to the dataframe seems to reapply this lost data. The 0 key is automatically generated when I run the df1_DateTimes, I think it just makes the column header titled 0.
All of the input datetime data is in the format dd/mm/yyyy HH:MM. However, when I make this "df1_DateTimes", the datetimes are mm/dd/yyyy HH:MM. And the column length is equal to that of the data before it was resampled.
I'm wondering if anyone knows of a way to make this "df1_DateTimes" in the format dd/mm/yyyy HH:MM, and to have the length of the column to be the same length of the resampled data? The latter isn't as important because I could just have a bunch of empty data. I've tried things like putting format='%d%m%y %H:%M', but it wasn't seeming to work.
Or if anyone knows how to resample the data and not lose the DateTimes? And have the DateTimes in 1 min increments as well? Any information on any of this would be greatly appreciated. Just as long as the end result is a dataframe with the values resampled to every minute, and the DateTime column intact, with the datatype of the DateTime column to be datetime64 (so I can merge it with my other datasets). I have included my code below.
df1 = pd.read_csv('PATH',
parse_dates=True, usecols=[0,7,10,13,28],
infer_datetime_format=True, index_col='DateTime')
# Resample data to take minute averages
df1.dropna(inplace=True) # Drops missing values
df1=(df1.resample('Min').mean())
df1.to_csv('df1', index=False, encoding='utf-8-sig')
df1_DateTimes = pd.to_datetime(df1.index.values)
df1_DateTimes = df1_DateTimes.to_frame()
df1_DateTimes.to_csv('df1_DateTimes', index=False, encoding='utf-8-sig'`
Thanks for reading and hope to hear back.
import datetime
df1__DateTimes = k
k['TITLE OF DATES COLUMN'] = k['TITLES OF DATES COLUMN'].datetime.strftime('%d/%m/%y')
I think using the above snippet solves your issue.
It assigns the date column to the formatted version (dd/mm/yy) of itself.
More on the Kite docs
I am trying to change the format of the date in a pandas dataframe.
If I check the date in the beginning, I have:
df['Date'][0]
Out[158]: '01/02/2008'
Then, I use:
df['Date'] = pd.to_datetime(df['Date']).dt.date
To change the format to
df['Date'][0]
Out[157]: datetime.date(2008, 1, 2)
However, this takes a veeeery long time, since my dataframe has millions of rows.
All I want to do is change the date format from MM-DD-YYYY to YYYY-MM-DD.
How can I do it in a faster way?
You should first collapse by Date using the groupby method to reduce the dimensionality of the problem.
Then you parse the dates into the new format and merge the results back into the original DataFrame.
This requires some time because of the merging, but it takes advantage from the fact that many dates are repeated a large number of times. You want to convert each date only once!
You can use the following code:
date_parser = lambda x: pd.datetime.strptime(str(x), '%m/%d/%Y')
df['date_index'] = df['Date']
dates = df.groupby(['date_index']).first()['Date'].apply(date_parser)
df = df.set_index([ 'date_index' ])
df['New Date'] = dates
df = df.reset_index()
df.head()
In my case, the execution time for a DataFrame with 3 million lines reduced from 30 seconds to about 1.5 seconds.
I'm not sure if this will help with the performance issue, as I haven't tested with a dataset of your size, but at least in theory, this should help. Pandas has a built in parameter you can use to specify that it should load a column as a date or datetime field. See the parse_dates parameter in the pandas documentation.
Simply pass in a list of columns that you want to be parsed as a date and pandas will convert the columns for you when creating the DataFrame. Then, you won't have to worry about looping back through the dataframe and attempting the conversion after.
import pandas as pd
df = pd.read_csv('test.csv', parse_dates=[0,2])
The above example would try to parse the 1st and 3rd (zero-based) columns as dates.
The type of each resulting column value will be a pandas timestamp and you can then use pandas to print this out however you'd like when working with the dataframe.
Following a lead at #pygo's comment, I found that my mistake was to try to read the data as
df['Date'] = pd.to_datetime(df['Date']).dt.date
This would be, as this answer explains:
This is because pandas falls back to dateutil.parser.parse for parsing the strings when it has a non-default format or when no format string is supplied (this is much more flexible, but also slower).
As you have shown above, you can improve the performance by supplying a format string to to_datetime. Or another option is to use infer_datetime_format=True
When using any of the date parsers from the answers above, we go into the for loop. Also, when specifying the format we want (instead of the format we have) in the pd.to_datetime, we also go into the for loop.
Hence, instead of doing
df['Date'] = pd.to_datetime(df['Date'],format='%Y-%m-%d')
or
df['Date'] = pd.to_datetime(df['Date']).dt.date
we should do
df['Date'] = pd.to_datetime(df['Date'],format='%m/%d/%Y').dt.date
By supplying the current format of the data, it is read really fast into datetime format. Then, using .dt.date, it is fast to change it to the new format without the parser.
Thank you to everyone who helped!
I have a pandas dataframe column that pandas currently thinks is an object. It's written as "58:42.5" which is the minutes then seconds and fractions of a second. I want to convert that to a time type so that I can subtract the two time columns I have to get a duration.
I've tried:
merged['started_at'] = pd.to_datetime(merged['started_at'],format='%M:%S').dt.time
However, that returns an error saying that the .5 was not converted. (ValueError: unconverted data remains: .5). I can I convert my entire two columns of time data to the correct format so that I can eventually do math with them.
My columns look like:
started_at , ended_at
58:42.5 , 00:02.3
00:55.5 , 02:13.9
Thanks for your help.
You can convert to timedelta without specifying a format:
x = '58:42.5'
res = pd.to_timedelta('00:'+x)
Timedelta('0 days 00:58:42.500000')
This transformation can easily be applied to a series:
merged['started_at'] = pd.to_timedelta('00:' + merged['started_at'], errors='coerce')
Your subsequent operations, for example difference between timedelta objects, should follow naturally.
I have a column in my dataframe that lists time in HH:MM:SS. When I run dtype on the column, it comes up with dtype('o') and I want to be able to use it as the x-axis for plotting some of my other signals. I saw previous documentation on using to_datetime and tried to use that to convert it to a usable time format for matplotlib.
Used pandas version is 0.18.1
I used:
time=pd.to_datetime(df.Time,format='%H:%M:%S')
where the output then becomes:
time
0 1900-01-01 00:00:01
and is carried out for the rest of the data points in the column.
Even though I specified just hour,minutes,and seconds I am still getting date. Why is that? I also tried
time.hour()
just to extract the hour portion but then I get an error that it doesn't have an 'hour' attribute.
Any help is much appreciated! Thanks!
Now in 2019, using pandas 0.25.0 and Python 3.7.3.
(Note : Edited answer to take plotting in account)
Even though I specified just hour,minutes,and seconds I am still getting date. Why is that?
According to pandas documentation I think it's because in a pandas Timestamp (equivalent of Datetime) object, the arguments year, month and day are mandatory, while hour, minutes and seconds are optional.
Therefore if you convert your object-type object in a Datetime, it must have a year-month-day part - if you don't indicate one, it will be the default 1900-01-01.
Since you also have a Date column in your sample, you can use it to have a datetime column with the right dates that you can use to plot :
import pandas as pd
df['Time'] = df.Date + " " + df.Time
df['Time'] = pd.to_datetime(df['Time'], format='%m/%d/%Y %H:%M:%S')
df.plot('Time', subplots=True)
With this your 'Time' column will display values like : 2016-07-25 01:12:07 and its dtype is datetime64[ns].
That being said, IF you plot day by day and you only want to compare times within a day (and not dates+times), having a default date does not seem bothering as long as it's the same date for all times - the times will be correctly compared on a same day, be it a wrong one.
And in the least likely case you would still want a time-only column, this is the reverse operation :
import pandas as pd
df['Time-only'] = pd.to_datetime(df['Time'], format='%H:%M:%S').dt.time
As explained before, it doesn't have a date (year-month-day) so it cannot be a datetime object, therefore this column will be in Object format.
You can extract a time object like:
import pandas as pd
df = pd.DataFrame([['12:10:20']], columns={"time": "item"})
time = pd.to_datetime(df.time, format='%H:%M:%S').dt.time[0]
After which you can extract desired properties as:
hour = time.hour
(Source)
How can I drop rows from Dataframe df if the dates associated with df['maturity_dt'] are less that today's date?
I am currently doing the following:
todays_date = datetime.date.today()
datenow = datetime.datetime.combine(todays_date, datetime.datetime.min.time()) #Converting to datetime
for (i,row) in df.iterrows():
if datetime.datetime.strptime(row['maturity_dt'], '%Y-%m-%d %H:%M:%S.%f') < datenow):
df.drop(df.index[i])
However, its taking too long and I was hoping to do something like: df = df[datetime.datetime.strptime(df['maturity_dt'], '%Y-%m-%d %H:%M:%S.%f') < datenow, but this results in the error TypeError: must be str, not Series
Thank You
Haven't tried it but maybe the pandas native functions will iterate faster. Something like:
df['dt']=pandas.Datetimeindex(df['maturity_dt'])
newdf=df.loc[df['dt']<=todays_date].copy()
Instead of parsing the date in each row, you could format your comparison date in the same format as these dates are stored and then you could just do a string comparison.
Also, if there is a way to drop multiple rows in a single call, you could use your loop just to gather the indices of those rows to be dropped, then use that call to drop them in bunches.