TypeError: sequence item 0: expected str instance, int found?

Asked Nov 07 '19 at 15:45

Active Nov 07 '19 at 15:48

Viewed 1,471 times

print ('\n'.join([' '.join([i for i in range(1,j)]) for j in range(2,6)]))

Error

TypeError: sequence item 0: expected str instance, int found

I went through the link TypeError: sequence item 0: expected string, int found telling string.join connects elements inside list of strings, not ints.

I have updated my code like print ('\n'.join([' '.join([str(i) for i in range(1,j)]) for j in range(2,6)]))

I got the error TypeError: 'str' object is not callable

Following code is working fine

print ('\n'.join([' '.join(['i' for i in range(1,j)]) for j in range(2,6)]))

My Out

i
i i
i i i
i i i i

asked Nov 07 '19 at 15:45

sre

`print ('\n'.join([' '.join([str(i) for i in range(1,j)]) for j in range(2,6)]))` works for me with no errors – Chris Doyle Nov 07 '19 at 15:48
`print ('\n'.join([' '.join([str(i) for i in range(1,j)]) for j in range(2,6)]))` doesn't give me any errors – Hippolippo Nov 07 '19 at 15:48
@ChrisDoyle do `str = 'a string'` and try again – DeepSpace Nov 07 '19 at 15:50
@Hippolippo Because you did not assign a string to `str` like OP has – DeepSpace Nov 07 '19 at 15:51
1

@DeepSpace ah you think in some previous part of the code he has overwritten the str function by allocating an actual string to it? – Chris Doyle Nov 07 '19 at 15:51
2

@ChrisDoyle OP **did** assign something to `str` – DeepSpace Nov 07 '19 at 15:52
1

Just don't assign variables to keywords ever and you will be fine – Hippolippo Nov 07 '19 at 15:53
You can make your first line of code be `to_string = str` and then just call `to_string` instead of `str` – Hippolippo Nov 07 '19 at 15:55

0 Answers0