CODEWITHSUNDEEP
pythonstring
aikipooh
aikipooh

Reputation: 235

Printing a float with the least possible digits after decimal point

Is there an elegant way to have a float→str output with eg zero or one digits?

8.5 should become '8.5', and 9.0 should become '9'.

Update. Well, to make it more precise, the way, that'll work for any float, and will output one digit save for the "integer" floats, which should be formatted as integers, id est without decimal point.

Upvotes: 0

Views: 1564

Answers (4)

aikipooh
aikipooh

Reputation: 235

Turned out that for my use case I can use Decimal instead of float, setting it from strings, and it does the right thing™:

>>> 'value={}'.format(Decimal('5'))
'value=5'
>>> 'value={}'.format(Decimal('5.5'))
'value=5.5'

Upvotes: 1

jpp
jpp

Reputation: 164633

With Python 3.6+ you should consider formatted string literals (PEP 498):

x = 8.5
y = 9.0
z = 8.05

for i in (x, y, z):
    print(f'{i:.9g}')

8.5
9
8.05

Manually, you can define a function to compare float and int values:

def make_str(x):
    float_val = float(x)
    int_val = int(float_val)
    if float_val == int_val:
        return str(int_val)
    return str(float_val)

make_str(9.0)  # '9'
make_str(8.5)  # '8.5'

Upvotes: 3

Krishna
Krishna

Reputation: 1362

Try str.format with "General format" as described here.

In [1]: print('{:.9g}'.format(8.5))
8.5

In [2]: print('{:.9g}'.format(9.0))
9

Upvotes: 3

Austin
Austin

Reputation: 26039

Just a replace:

>>> s = 8.5
>>> str(s).replace('.0', '')
8.5
>>> s = 9.0
>>> str(s).replace('.0', '')
9

Cases like 8.05 fail using replace. To handle such cases, if you want, use re.sub:

import re

s = 8.05
print(re.sub(r'\.0*$', '', str(s)))
# 8.05

Upvotes: 0

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