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Answers in Scripting and Automation | By Some Guy


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Tags: Python, Scripting, Errors, Programming, TypeError

Profile Photo - Some GuySome GuyBasic Subscriber

I have an issue with my code in Python. I keep getting an error about the variable type: "TypeError: 'float' object cannot be interpreted as an integer". I do understand what the code says but I have no idea how to resolve it. Any help is appreciated

Download more RAM. 🐏 ⨉ 0Posted by Some Guy 2 years ago 🕓 Posted at 13 January, 2019 05:40 AM PST

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Profile Photo - August R. GarciaAugust R. GarciaLARPing as a Sysadmi...Portland, ORSite Owner

One possible reason why you're running into this error is because of a float being divided to a cleanly-rounded number, but that result still being a float. This happens with the range() function, for example.

>>> x = 4.0
>>> y = x / 2
>>> print x
>>> print y
>>> for i in range (y):
...     print "y loop"
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: range() integer end argument expected, got float.

In this scenario, you can cast the float to an integer. Note that there are generally better options than typecasting values that will prevent errors from showing up further down the line:

>>> for i in range ( int(y) ):
...     print "y loop"
y loop
y loop

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Posted by August R. Garcia 2 years ago 🕓 Posted at 31 January, 2019 20:42 PM PST

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Profile Photo - rahuldevrahuldev

When we try to use float numbers in Python range() function, we get a type error 'float' object that cannot be interpreted as an integer. 

One option is is to convert float to int and use it in range. But that is not the ideal way. So you should use write custom range() function that will allow float arguments. you can use NumPy’s arange() and linspace() functions to generate the range of float numbers. Refer to this range for float numbers


import numpy
x = 2.0
y = 10.5
for i in numpy.arange(x, y):
    print(i, end=', ')

Download more RAM. 🐏 ⨉ 0Posted by rahuldev 1 year ago 🕓 Posted at 24 March, 2020 04:03 AM PDT

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