How to turn a float number like 293.4662543 into 293.47 in python?

Question:

How to shorten the float result I got? I only need 2 digits after the dot. Sorry I really don’t know how to explain this better in English…

Thanks

Asked By: Shane

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Answers:

>>> print "%.2f" % 293.44612345
293.45
Answered By: ghostdog74

From :
Python Docs
round(x[, n])
Return the floating point value x rounded to n digits after the decimal point. If n is omitted, it defaults to zero. The result is a floating point number. Values are rounded to the closest multiple of 10 to the power minus n; if two multiples are equally close, rounding is done away from 0 (so. for example, round(0.5) is 1.0 and round(-0.5) is -1.0).

Note The behavior of round() for floats can be surprising: for example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This is not a bug: it’s a result of the fact that most decimal fractions can’t be represented exactly as a float. See Floating Point Arithmetic: Issues and Limitations for more information.

Looks like round (293.466….[, 2]) would do it,

Answered By: onaclov2000

x = round(293.4662543, 2)

Answered By: Ethan Shepherd

If you want a number, use the round() function:

>>> round(12.3456, 2)
12.35

(but +1 for Michael’s answer re. the Decimal type.)

If you want a string:

>>> print "%.2f" % 12.34567
12.35
Answered By: RichieHindle

From The Floating-Point Guide’s Python cheat sheet:

"%.2f" % 1.2399 # returns "1.24"
"%.3f" % 1.2399 # returns "1.240"
"%.2f" % 1.2 # returns "1.20"

Using round() is the wrong thing to do, because floats are binary fractions which cannot represent decimal digits accurately.

If you need to do calculations with decimal digits, use the Decimal type in the decimal module.

Answered By: Michael Borgwardt

If you need numbers like 2.3k or 12M, this function does the job:

def get_shortened_integer(number_to_shorten):
    """ Takes integer and returns a formatted string """
    trailing_zeros = floor(log10(abs(number_to_shorten)))
    if trailing_zeros < 3:
        # Ignore everything below 1000
        return trailing_zeros
    elif 3 <= trailing_zeros <= 5:
        # Truncate thousands, e.g. 1.3k
        return str(round(number_to_shorten/(10**3), 1)) + 'k'
    elif 6 <= trailing_zeros <= 8:
        # Truncate millions like 3.2M
        return str(round(number_to_shorten/(10**6), 1)) + 'M'
    else:
        raise ValueError('Values larger or equal to a billion not supported')

Results:

>>> get_shortened_integer(2300)
2.3k # <-- str

>>> get_shortened_integer(1300000)
1.3M # <-- str
Answered By: chribsen

I hope this will help.

def do(*args):
    formattedList = [float("{:.2f}".format(num)) for num in args]
    _result =(sum(formattedList))
    result = round(_result,2)
    return result


print(do(23.2332,45.24567,67,54.27))

Result:

189.75
Answered By: Xavi Martínez

One way:

>>> number = 1
>>> '{:.2f}'.format(number) #1.00
>>> '{:.3f}'.format(number) #1.000

second way:

>>> '%.2f' % number #1.00
>>> '%.3f' % number #1.000

see "format python"

Answered By: Alexey Nikonov
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