I have a data frame with only dates in one column.
My objective is to calculate the number of days between each date, in order to get the average number of days between two dates (each date corresponds to an operation).
I tried doing something like this :
for i in range(len(df)):
if i != 0:
d0 = df.iloc[[i-1,1]]
d1 = df.iloc[[i,1]]
L.append((d1 - d0).days)
But got the error message : 'unsupported operand type(s) for -: 'str' and 'str''
You can subtract a date from another if they're in proper format. Maybe a timestamp in seconds, maybe a proper datetime object. As you might've noticed, you can't subtract strings.
If the date is in the ISO format, it's the easiest, just do this before the subtraction:
from datetime import datetime
d0 = datetime.fromisoformat(d0)
d1 = datetime.fromisoformat(d1)
The result will be in a datetime.timedelta format with a .total_seconds() method and even a .days attribute, so you get days like this:
difference_in_days = (d1 - d0).days
If it's something else than an ISO string, check out this question:
Converting string into datetime
I have a dataframe with timestamp of BirthDate = 2001-10-10 11:01:04.343
How can I get an actual age?
I tried like that:
i.loc[0, "BirthDate"] = pd.to_datetime('today').normalize() - i.loc[0, "BirthDate"].normalize()
output is: 7248 days 00:00:00
but is there any better method which give me just output 19 years?
If i use:
(i.loc[0, "BirthDate"] = pd.to_datetime('today').normalize() - i.loc[0, "BirthDate"].normalize())/365
the output is:
19 days 20:34:50:958904109 and it is type <class 'pandas.timedeltas.Timedelta>
The timedelta result is wrong because you are dividing by 365 where you shouldn't. It actually means 19.86 years.
In some more detail, you are taking a value which is in years, and dividing it with 365; so now you have a result which shows 1/365 of the input duration. The proper way to get the result in years is to divide by another timedelta.
>>> from datetime import timedelta
>>> t = timedelta(days=7248)
>>> 7248/365.0
19.85753424657534
>>> print(t)
7248 days, 0:00:00
>>> t/timedelta(days=365)
19.85753424657534
>>> # years
How exactly to represent a year is not completely well-defined. You could use timedelta(days=365.2425) for the arithmetically correct length of a year, but then of course that produces odd results if you try to convert that back to a resolution where hours and minutes are important.
First, delete the last part of the timestamp and then the following python code can be applied:
from datetime import datetime, date
def calculate_age(born):
born = datetime.strptime(born, "%d/%m/%Y").date()
today = date.today()
return today.year - born.year - ((today.month, today.day) < (born.month, born.day))
df['Age'] = df['Date of birth'].apply(calculate_age)
print(df)
Both types of data have been converted into dates, and it still is telling me it can't subtract them. I've watched many tutorials and looked at stack overflow for hours, but all the solutions say that the only possible problem is an incorrect form of data which is not true. The error message is :
>unsupported operand type(s) for -: 'datetime.datetime' and 'datetime.date'<
The text file contains only this:
Birthday 30/5/2021
def display():
with open ('countdown.txt', 'r') as file:
for line in file:
data = line.split(' ')
title = data[0]
date_str = data[1]
date = datetime.datetime.strptime(date_str, '%d/%m/%y')
time_between = datetime.date.today()-date
print(title + 'is' + time_between + 'days away!')
display()
One value is date+time and one is date only, they do not match. Replace datetime.date.today() with datetime.datetime.today().
The result will be a datetime.timedelta object.
There are two different date methods, one in datetime and one in date:
import datetime
dt = datetime.datetime.today()
print(dt, type(dt))
# 2020-12-10 20:45:20.661116 <class 'datetime.datetime'>
d = datetime.date.today()
print(d, type(d))
# 2020-12-10 <class 'datetime.date'>
As your date is a datetime object, you need to subtract it with another datetime, so you have to use:
time_between = datetime.datetime.today() - date
I would like to write a function that takes a date entered by the user, stores it with the shelve function and prints the date thirty days later when called.
I'm trying to start with something simple like:
import datetime
def getdate():
date1 = input(datetime.date)
return date1
getdate()
print(date1)
This obviously doesn't work.
I've used the answers to the above question and now have that section of my program working! Thanks!
Now for the next part:
I'm trying to write a simple program that takes the date the way you instructed me to get it and adds 30 days.
import datetime
from datetime import timedelta
d = datetime.date(2013, 1, 1)
print(d)
year, month, day = map(int, d.split('-'))
d = datetime.date(year, month, day)
d = dplanted.strftime('%m/%d/%Y')
d = datetime.date(d)+timedelta(days=30)
print(d)
This gives me an error:
year, month, day = map(int, d.split('-'))
AttributeError: 'datetime.date' object has no attribute 'split'
Ultimately what I want is have 01/01/2013 + 30 days and print 01/30/2013.
Thanks in advance!
The input() method can only take text from the terminal. You'll thus have to figure out a way to parse that text and turn it into a date.
You could go about that in two different ways:
Ask the user to enter the 3 parts of a date separately, so call input() three times, turn the results into integers, and build a date:
year = int(input('Enter a year'))
month = int(input('Enter a month'))
day = int(input('Enter a day'))
date1 = datetime.date(year, month, day)
Ask the user to enter the date in a specific format, then turn that format into the three numbers for year, month and day:
date_entry = input('Enter a date in YYYY-MM-DD format')
year, month, day = map(int, date_entry.split('-'))
date1 = datetime.date(year, month, day)
Both these approaches are examples; no error handling has been included for example, you'll need to read up on Python exception handling to figure that out for yourself. :-)
Thanks. I have been trying to figure out how to add info to datetime.datetime(xxx) and this explains it nicely. It's as follows
datetime.datetime(year,month, day, hour, minute, second) with parameters all integer. It works!
Use the dateutils module
from dateutil import parser
date = parser.parse(input("Enter date: "))
you can also use
import datetime
time_str = input("enter time in this format yyyy-mm-dd")
time=datetime.datetime.strptime(time_str, "%Y-%m-%d")
datetime.datetime.strptime() strips the given string in the format you give it.
Check the library as
import datetime
and follow syntax
date = datetime.datetime(2013, 1, 1)
I have a .txt file data-set like this with the date column of interest:
1181206,3560076,2,01/03/2010,46,45,M,F
2754630,2831844,1,03/03/2010,56,50,M,F
3701022,3536017,1,04/03/2010,40,38,M,F
3786132,3776706,2,22/03/2010,54,48,M,F
1430789,3723506,1,04/05/2010,55,43,F,M
2824581,3091019,2,23/06/2010,59,58,M,F
4797641,4766769,1,04/08/2010,53,49,M,F
I want to work out the number of days between each date and 01/03/2010 and replace the date with the days offset {0, 2, 3, 21...} yielding an output like this:
1181206,3560076,2,0,46,45,M,F
2754630,2831844,1,2,56,50,M,F
3701022,3536017,1,3,40,38,M,F
3786132,3776706,2,21,54,48,M,F
1430789,3723506,1,64,55,43,F,M
2824581,3091019,2,114,59,58,M,F
4797641,4766769,1,156,53,49,M,F
I've been trying for ages and its getting really frustrating. I've tried converting to datetime using the datetime.datetime.strptime( '01/03/2010', "%d/%m/%Y").date() method and then subtracting the two dates but it gives me an output of e.g. '3 days, 0:00:00' but I can't seem to get an output of only the number!
The difference between two dates is a timedelta. Any timedelta instance has days attribute that is an integer value you want.
This is fairly simple. Using the code you gave:
date1 = datetime.datetime.strptime('01/03/2010', '%d/%m/%Y').date()
date2 = datetime.datetime.strptime('04/03/2010', '%d/%m/%Y').date()
You get two datetime objects.
(date2-date1)
will give you the time delta. The mistake you're making is to convert that timedelta to a string. timedelta objects have a days attribute. Therefore, you can get the number of days using it:
(date2-date1).days
This generates the desired output.
Using your input (a bit verbose...)
#!/usr/bin/env python
import datetime
with open('input') as fd:
d_first = datetime.date(2010, 03, 01)
for line in fd:
date=line.split(',')[3]
day, month, year= date.split(r'/')
d = datetime.date(int(year), int(month), int(day))
diff=d - d_first
print diff.days
Gives
0
2
3
21
64
114
156
Have a look at pleac, a lot of date-example there using python.