TypeError: 'str' object cannot be interpreted as an integer python for loop

Question

My function should return data starting from a given year and given month to today's year and month. I am getting error TypeError: 'str' object cannot be interpreted as an integer

Defining function

def frequency_database(df,start_month,start_year):
    data = pd.DataFrame([])
    import datetime
    start_month=int(start_month)
    start_year=int(start_year)
    today = datetime.date.today()
    today_year=today.strftime('%Y')

    for y in range(start_year,today_year):
        print('Outer loop enter for year', y)
        # Some function...
        for x in range(start_month,13):
            print('Inner loop enter for month number',x)
            # Some function...

Calling funtion

frequency_database(df,1,2015)

Error

TypeError: 'str' object cannot be interpreted as an integer

Stack Trace as requested

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-37-06567ae5d027> in <module>
     12             print('Inner loop enter for month number',x)
     13 
---> 14 frequency_database(df,1,2015)

<ipython-input-37-06567ae5d027> in frequency_database(df, start_month, start_year)
      7     today_year=today.strftime('%Y')
      8 
----> 9     for y in range(start_year,today_year):
     10         print('Outer loop enter for year', y)
     11         for x in range(start_month,13):

TypeError: 'str' object cannot be interpreted as an integer

Replace today_year=today.strftime('%Y') with today_year=int(today.strftime('%Y')) — han solo
– han solo, Commented Jan 3, 2020 at 9:14
Please include the stack trace of your exception and tell us which line it is indicating — khelwood
– khelwood, Commented Jan 3, 2020 at 9:14

han solo · Accepted Answer · 2020-01-03 09:32:05Z

5

The problem is you tried to give range a str which is today_year=today.strftime('%Y').

Just replace the line

today_year=today.strftime('%Y')

with

today_year=int(today.strftime('%Y'))

As pointed out by Stargazer, you could do,

today_year = today.year

instead of converting the str to int

edited Jan 3, 2020 at 9:32

answered Jan 3, 2020 at 9:15

han solo

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1 Comment

Stargazer Over a year ago

No need to that. Just today.year will return the year as an int

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