Any ideas on how I can manipulate my current date-time data to make it suitable for use when converting the datatype to time?
For example:
df1['Date/Time'] = pd.to_datetime(df1['Date/Time'])
The current format for the data is mm/dd 00:00:00
an example of the column in the dataframe can be seen below.
Date/Time Dry_Temp[C] Wet_Temp[C] Solar_Diffuse_Rate[[W/m2]] \
0 01/01 00:10:00 8.45 8.237306 0.0
1 01/01 00:20:00 7.30 6.968360 0.0
2 01/01 00:30:00 6.15 5.710239 0.0
3 01/01 00:40:00 5.00 4.462898 0.0
4 01/01 00:50:00 3.85 3.226244 0.0
For the condition where the hour is denoted as 24, you have two choices. First you can simply reset the hour to 00 and second you can reset the hour to 00 and also add 1 to the date.
In either case the first step is detecting the condition which can be done with a simple find statement t.find(' 24:')
Having detected the condition in the first case it is a simple matter of reseting the hour to 00 and proceeding with the process of formatting the field. In the second case, however, adding 1 to the day is a little more complicated because of the fact you can roll over to next month.
Here is the approach I would use:
Given a df of form:
Date Time
0 01/01 00:00:00
1 01/01 00:24:00
2 01/01 24:00:00
3 01/31 24:00:00
The First Case
def parseDate2(tx):
ti = tx.find(' 24:')
if ti >= 0:
tk = pd.to_datetime(tx[:5]+' 00:'+tx[10:], format= '%m/%d %H:%M:%S')
return tk + du.relativedelta.relativedelta(hours=+24)
return pd.to_datetime(tx, format= '%m/%d %H:%M:%S')
df['Date Time'] = df['Date Time'].apply(lambda x: parseDate(x))
Produces the following:
Date Time
0 1900-01-01 00:00:00
1 1900-01-01 00:24:00
2 1900-01-01 00:00:00
3 1900-01-31 00:00:00
For the second case, I employed the dateutil relativedelta library and slightly modified my parseDate funstion as shown below:
import dateutil as du
def parseDate2(tx):
ti = tx.find(' 24:')
if ti >= 0:
tk = pd.to_datetime(tx[:5]+' 00:'+tx[10:], format= '%m/%d %H:%M:%S')
return tk + du.relativedelta.relativedelta(hours=+24)
return pd.to_datetime(tx, format= '%m/%d %H:%M:%S')
df['Date Time'] = df['Date Time'].apply(lambda x: parseDate2(x))
Yields:
Date Time
0 1900-01-01 00:00:00
1 1900-01-01 00:24:00
2 1900-01-02 00:00:00
3 1900-02-01 00:00:00
To access the values of the datetime (namely the time), you can use:
# These are now in a usable format
seconds = df1['Date/Time'].dt.second
minutes = df1['Date/Time'].dt.minute
hours = df1['Date/Time'].dt.hours
And if need be, you can create its own independent time series with:
df1['Dat/Time'].dt.time
Related
I am currently having issues with date-time format, particularly converting string input to the correct python datetime format
Date/Time Dry_Temp[C] Wet_Temp[C] Solar_Diffuse_Rate[[W/m2]] \
0 01/01 00:10:00 8.45 8.237306 0.0
1 01/01 00:20:00 7.30 6.968360 0.0
2 01/01 00:30:00 6.15 5.710239 0.0
3 01/01 00:40:00 5.00 4.462898 0.0
4 01/01 00:50:00 3.85 3.226244 0.0
These are current examples of timestamps I have in my time, I have tried splitting date and time such that I now have the following columns:
WC_Humidity[%] WC_Htgsetp[C] WC_Clgsetp[C] Date Time
0 55.553640 18 26 1900-01-01 00:10:00
1 54.204342 18 26 1900-01-01 00:20:00
2 51.896272 18 26 1900-01-01 00:30:00
3 49.007770 18 26 1900-01-01 00:40:00
4 45.825810 18 26 1900-01-01 00:50:00
I have managed to get the year into datetime format, but there are still 2 problems to resolve:
the data was not recorded in 1900, so I would like to change the year in the Date,
I get the following error whent rying to convert time into time datetime python format
pandas/_libs/tslibs/strptime.pyx in pandas._libs.tslibs.strptime.array_strptime()
ValueError: time data '00:00:00' does not match format ' %m/%d %H:%M:%S' (match)
I tried having 24:00:00, however, python didn't like that either...
preferences:
I would prefer if they were both in the same cell without having to split this information into two columns.
I would also like to get rid of the seconds data as the data was recorded in 10 min intervals so there is no need for seconds in my case.
Any help would be greatly appreciated.
the data was not recorded in 1900, so I would like to change the year in the Date,
datetime.datetime.replace method of datetime.datetime instance is used for this task consider following example:
import pandas as pd
df = pd.DataFrame({"when":pd.to_datetime(["1900-01-01","1900-02-02","1900-03-03"])})
df["when"] = df["when"].apply(lambda x:x.replace(year=2000))
print(df)
output
when
0 2000-01-01
1 2000-02-02
2 2000-03-03
Note that it can be used also without pandas for example
import datetime
d = datetime.datetime.strptime("","") # use all default values which result in midnight of Jan 1 of year 1900
print(d) # 1900-01-01 00:00:00
d = d.replace(year=2000)
print(d) # 2000-01-01 00:00:00
I have the following df:
time_series date sales
store_0090_item_85261507 1/2020 1,0
store_0090_item_85261501 2/2020 0,0
store_0090_item_85261500 3/2020 6,0
Being 'date' = Week/Year.
So, I tried use the following code:
df['date'] = df['date'].apply(lambda x: datetime.strptime(x + '/0', "%U/%Y/%w"))
But, return this df:
time_series date sales
store_0090_item_85261507 2020-01-05 1,0
store_0090_item_85261501 2020-01-12 0,0
store_0090_item_85261500 2020-01-19 6,0
But, the first day of the first week of 2020 is 2019-12-29, considering sunday as first day. How can I have the first day 2020-12-29 of the first week of 2020 and not 2020-01-05?
From the datetime module's documentation:
%U: Week number of the year (Sunday as the first day of the week) as a zero padded decimal number. All days in a new year preceding the first Sunday are considered to be in week 0.
Edit: My originals answer doesn't work for input 1/2023 and using ISO 8601 date values doesn't work for 1/2021, so I've edited this answer by adding a custom function
Here is a way with a custom function
import pandas as pd
from datetime import datetime, timedelta
##############################################
# to demonstrate issues with certain dates
print(datetime.strptime('0/2020/0', "%U/%Y/%w")) # 2019-12-29 00:00:00
print(datetime.strptime('1/2020/0', "%U/%Y/%w")) # 2020-01-05 00:00:00
print(datetime.strptime('0/2021/0', "%U/%Y/%w")) # 2020-12-27 00:00:00
print(datetime.strptime('1/2021/0', "%U/%Y/%w")) # 2021-01-03 00:00:00
print(datetime.strptime('0/2023/0', "%U/%Y/%w")) # 2023-01-01 00:00:00
print(datetime.strptime('1/2023/0', "%U/%Y/%w")) # 2023-01-01 00:00:00
#################################################
df = pd.DataFrame({'date':["1/2020", "2/2020", "3/2020", "1/2021", "2/2021", "1/2023", "2/2023"]})
print(df)
def get_first_day(date):
date0 = datetime.strptime('0/' + date.split('/')[1] + '/0', "%U/%Y/%w")
date1 = datetime.strptime('1/' + date.split('/')[1] + '/0', "%U/%Y/%w")
date = datetime.strptime(date + '/0', "%U/%Y/%w")
return date if date0 == date1 else date - timedelta(weeks=1)
df['new_date'] = df['date'].apply(lambda x:get_first_day(x))
print(df)
Input
date
0 1/2020
1 2/2020
2 3/2020
3 1/2021
4 2/2021
5 1/2023
6 2/2023
Output
date new_date
0 1/2020 2019-12-29
1 2/2020 2020-01-05
2 3/2020 2020-01-12
3 1/2021 2020-12-27
4 2/2021 2021-01-03
5 1/2023 2023-01-01
6 2/2023 2023-01-08
You'll want to use ISO week parsing directives, Ex:
import pandas as pd
date = pd.Series(["1/2020", "2/2020", "3/2020"])
pd.to_datetime(date+"/1", format="%V/%G/%u")
0 2019-12-30
1 2020-01-06
2 2020-01-13
dtype: datetime64[ns]
you can also shift by one day if the week should start on Sunday:
pd.to_datetime(date+"/1", format="%V/%G/%u") - pd.Timedelta('1d')
0 2019-12-29
1 2020-01-05
2 2020-01-12
dtype: datetime64[ns]
I have a pandas Dataframe in which one of the column is pandas datetime column created using pd.to_datetime()1. I want to extract the date and hour from each datetime object, in other words, I want to change the minute and seconds to 0.
I used normalize() to change the time to midnight but don't how how to change the time to start of the hour. Please suggest a way to do so.
making some test data and turning it into a dataframe
rng = pd.date_range('1/1/2018 11:59:00', periods=3, freq='min')
df = pd.DataFrame(rng)
print(df)
print(df[0].round('H'))
gives the input
0
0 2018-01-01 11:59:00
1 2018-01-01 12:00:00
2 2018-01-01 12:01:00
and rounded to the nearest hour gives
0
0 2018-01-01 12:00:00
1 2018-01-01 12:00:00
2 2018-01-01 12:00:00
and
print(df[0].dt.floor('H'))
gives
0
0 2018-01-01 11:00:00
1 2018-01-01 12:00:00
2 2018-01-01 12:00:00
if you always want to round down. Likewise, ceil('H') if you want to round up
I think you need to checkout pandas.Series.dt.strftime
Or try this:
import datetime
df=pd.DataFrame({'timestamp':[pd.Timestamp('today')]})
df['Date']=[pd.to_datetime(i.date())+ datetime.timedelta(hours=i.hour) for i in df['timestamp']]
I am working in a dataframe in Pandas that looks like this.
Identifier datetime
0 AL011851 00:00:00
1 AL011851 06:00:00
2 Al011851 12:00:00
This is my code so far:
import pandas as pd
hurricane_df = pd.read_csv("hurdat2.csv",parse_dates=['datetime'])
hurricane_df['datetime'] = pd.to_timedelta(hurricane_df['datetime'].dt.strftime('%H:%M:%S'))
hurricane_df
grouped = hurricane_df.groupby('datetime').size()
grouped
What I did was convert the datetime column to a timedelta to get the hours. I want to get the size of the datetime column but I want just hours like 1:00, 2:00, 3:00, etc. but I get minute intervals as well like 1:15 and 2:45.
Any way to just display the hour?
Thank you.
You can use pandas.Timestamp.round with Series.dt shortcut:
df['datetime'] = df['datetime'].dt.round('h')
So
... datetime
01:15:00
02:45:00
becomes
... datetime
01:00:00
03:00:00
df = pd.DataFrame({'Identifier':['AL011851','AL011851','AL011851'],'datetime': ["2018-12-08 16:35:23","2018-12-08 14:20:45", "2018-12-08 11:45:00"]})
df['datetime'] = pd.to_datetime(df['datetime'])
df
Identifier datetime
0 AL011851 2018-12-08 16:35:23
1 AL011851 2018-12-08 14:20:45
2 AL011851 2018-12-08 11:45:00
# Rounds to nearest hour
def roundHour(t):
return (t.replace(second=0, microsecond=0, minute=0, hour=t.hour)
+timedelta(hours=t.minute//30))
df.datetime=df.datetime.map(lambda t: roundHour(t)) # Step 1: Round to nearest hour
df.datetime=df.datetime.map(lambda t: t.strftime('%H:%M')) # Step 2: Remove seconds
df
Identifier datetime
0 AL011851 17:00
1 AL011851 14:00
2 AL011851 12:00
i have dataframe which contains 3 column for date and time: date ,depart time and arrive time. i want to make two columns of datetime (depart time and arrive time) using pandas so I use to_datetime function.
since the date column based only on the depart time, there are some cases where the depart time is around 23:00 and the arrive time is after 24:00 but the date stays the same. for instance:
depart datetime: 01/12/2017 23:58:00 arrive time 01/12/2017 00:30:00
how can i write a function that will update the day to the day after if the arrive time is after midnight? (in the example it should be arrive time 02/12/2017)
thanks
I think you can check difference is bellow 0 Timedelta and by mask add one day:
print (df)
depart time arrive time
0 01/12/2017 23:58:00 01/12/2017 00:30:00
1 01/12/2017 00:30:00 01/12/2017 23:58:00
df['depart time'] = pd.to_datetime(df['depart time'], dayfirst=True)
df['arrive time'] = pd.to_datetime(df['arrive time'], dayfirst=True)
m = (df['arrive time'] - df['depart time']) < pd.Timedelta(0)
Another condition should be:
m = (df['depart time'] - df['arrive time']).dt.days != -1
print (m)
0 True
1 False
dtype: bool
df['arrive time'] = df['arrive time'].mask(m, df['arrive time'] + pd.Timedelta(1, unit='d'))
print (df)
depart time arrive time
0 2017-12-01 23:58:00 2017-12-02 00:30:00
1 2017-12-01 00:30:00 2017-12-01 23:58:00