Converting strings into float and integers (if is necessary) in Python

Question

I've to call an API that sometimes returns me numerical values in string format, "288" instead of 288, or "0.1523" instead of 0.1513. Some other times, I get the proper numerical value, 39.

How can I make the function to format it to the proper value? This means:

If I get "288" convert it to an integer: 288.
If I get "0.323" convert it to a float: 0.323.
If I get 288 leave it as it is (its an integer already).
If I get 0.323 leave as it is (its a float already).

This is my try. The thing is that this also converts me all the floats into integers, and I don't want this. Can someone give me hand?

def parse_value(value):
    try:
       value = int(value)
    except ValueError:
        try:
            value = float(value)
        except ValueError:
            pass

Thanks in advance

Why are you making a distinction between float and int? For most usecases float(value) is enough. — MegaIng
– MegaIng, Commented Jun 30, 2020 at 14:16

Umair Mubeen · Accepted Answer · 2020-06-30 14:22:38Z

1

def parse_value(value):
    try:
       print(type(value))
       if type(value) is float:
           print(value)
       elif type(value) is int:
           print(value)
       elif type(value) is str:
           value = float(value)
           print(value)

    except ValueError:
        try:
           print(value)
        except ValueError:
            pass

answered Jun 30, 2020 at 14:22

Umair Mubeen

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Comments

Sanket Singh · Accepted Answer · 2020-06-30 14:21:20Z

1

Use following:

def convert(self, value)
    a=value
    b=float(a)
    if(b==int(b)):b=int(b) 
    
    print(b)

answered Jun 30, 2020 at 14:21

Sanket Singh

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Comments

Presbitero · Accepted Answer · 2020-06-30 14:23:33Z

0

Would this work for you?

def parse_value(value):
    f = float(value)
    r = f if f != int(f) else int(f)
    return r

answered Jun 30, 2020 at 14:23

Presbitero

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