Yield CSV row as list of complex list comprehension with if statements

Question

I am trying to yield each row of CSV as list of int, float, str respectively while using list comprehension with if statements.

How do I write a list comprehension with if statements that will convert each element of CSV row into int, float, str respectively ?

Problem: float created as string.

import csv

FILE = "file.csv"

def all_values(file):
    with open(file, "r") as file_open:
        reader = csv.reader(file_open)
        for row in reader:
            yield [
                    int(x)
                    if x.isnumeric() else float(x)
                    if x.isdecimal() else x
                    for x in row
            ]

values = [x for x in all_values(FILE)]

print(values)

Problem in part of code:

            yield [
                    int(x)
                    if x.isnumeric() else float(x)
                    if x.isdecimal() else x
                    for x in row
            ]

File:

100,"Jin",10.56
200,"Tonic",15.99
300,"Vodka",20.39

Output:

[[100, 'Jin', '10.56'], [200, 'Tonic', '15.99'], [300, 'Vodka', '20.39']]

Wrong type of values:

'10.56', '15.99', '20.39'

Thank you for your time.

Answer 1

As mentioned in the comments isdecimal() is returning False because . is not considered a decimal character. Instead you could change your code to be:

yield [
    int(x)
    if x.isnumeric() else float(x)
    if x.replace('.','',1).isdigit() else x
    for x in row
]

In this case the order of the checks matter, int must be first.