What is a clean way to convert a string percent to a float?

Question

I have looked in the standard library and on StackOverflow, and have not found a similar question. So, is there a way to do the following without rolling my own function? Bonus points if someone writes a beautiful function if there is no built in way.

def stringPercentToFloat(stringPercent)
    # ???
    return floatPercent

p1 = "99%"
p2 = "99.5%"
print stringPercentToFloat(p1)
print stringPercentToFloat(p2)

>>>> 0.99
>>>> 0.995

Asked By: Wulfram

||

Source

Answer 1

Use strip('%') , as:

In [9]: "99.5%".strip('%')
Out[9]: '99.5'             #convert this to float using float() and divide by 100

In [10]: def p2f(x):
   ....:    return float(x.strip('%'))/100
   ....: 

In [12]: p2f("99%")
Out[12]: 0.98999999999999999

In [13]: p2f("99.5%")
Out[13]: 0.995

Answered By: Ashwini Chaudhary

Answer 2

float(stringPercent.strip('%')) / 100.0

Answered By: Mark Ransom

Answer 3

Another way:
float(stringPercent[:-1]) / 100

Answered By: WKPlus

Answer 4

I wrote the following method that should always return the output to the exact same accuracy as the input, with no floating point errors such as in the other answers.

def percent_to_float(s):
    s = str(float(s.rstrip("%")))
    i = s.find(".")
    if i == -1:
        return int(s) / 100
    if s.startswith("-"):
        return -percent_to_float(s.lstrip("-"))
    s = s.replace(".", "")
    i -= 2
    if i < 0:
        return float("." + "0" * abs(i) + s)
    else:
        return float(s[:i] + "." + s[i:])

Explanation

Strip the “%” from the end.
If percent has no “.”, simply return it divided by 100.
If percent is negative, strip the “-” and re-call function, then convert the result back to a negative and return it.
Remove the decimal place.
Decrement i (the index the decimal place was at) by 2, because we want to shift the decimal place 2 spaces to the left.
If i is negative, then we need to pad with zeros.
- Example: Suppose the input is “1.33%”. To be able to shift the decimal place 2 spaces to the left, we would need to pad with a zero.
Convert to a float.

Test case (Try it online):

from unittest.case import TestCase

class ParsePercentCase(TestCase):
    tests = {
        "150%"              : 1.5,
        "100%"              : 1,
        "99%"               : 0.99,
        "99.999%"           : 0.99999,
        "99.5%"             : 0.995,
        "95%"               : 0.95,
        "90%"               : 0.9,
        "50%"               : 0.5,
        "66.666%"           : 0.66666,
        "42%"               : 0.42,
        "20.5%"             : 0.205,
        "20%"               : 0.2,
        "10%"               : 0.1,
        "3.141592653589793%": 0.03141592653589793,
        "1%"                : 0.01,
        "0.1%"              : 0.001,
        "0.01%"             : 0.0001,
        "0%"                : 0,
    }
    tests = sorted(tests.items(), key=lambda x: -x[1])

    def test_parse_percent(self):
        for percent_str, expected in self.tests:
            parsed = percent_to_float(percent_str)
            self.assertEqual(expected, parsed, percent_str)

    def test_parse_percent_negative(self):
        negative_tests = [("-" + s, -f) for s, f in self.tests]
        for percent_str, expected in negative_tests:
            parsed = percent_to_float(percent_str)
            self.assertEqual(expected, parsed, percent_str)

Answered By: Hat

Answer 5

Based on the answer from @WKPlus this solution takes into account the locales where the decimal point is either a point . or a comma ,

float("-3,5%".replace(',','.')[:-1]) / 100

Answered By: krema

Answer 6

Simply replace the % by e-2 before parsing to float :

float("99.5%".replace('%', 'e-2'))

It’s safer since the result will still be correct if there’s no % used.

Answered By: Florent B.

What is a clean way to convert a string percent to a float?

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