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Home/ Questions/Q 1079627
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T21:53:59+00:00 2026-05-16T21:53:59+00:00

in C#, I can do (#.####), which prints up to 4 significant digits after

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in C#, I can do (#.####), which prints up to 4 significant digits after the decimal point.
1.2 -> 1.2
1.234 -> 1.234
1.23456789 -> 1.2345

Afaik, in python, there is only the c-style %.4f which will always print to 4 decimal points padded with 0s at the end if needed.
I don’t want those 0s.

Any suggestions for what is the cleanest way to achieve what I need?

One possible solution is to print it first and trim ending 0s myself, but hoping to find more clever ways.

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  1. Editorial Team
    Editorial Team
    2026-05-16T21:54:00+00:00Added an answer on May 16, 2026 at 9:54 pm

    It looks like you can use the “g” format specifier for the .format() string function (only available in python 2.6+):

    >>> x = 1.2
    >>> print "{0:0.4g}".format(x)
    1.2
    >>> x = 1.234565789
    >>> print "{0:0.4g}".format(x)
    1.235
    

    This includes digits before and after the decimal point, which is why you get 1.235 (rounded 1.2345 to 4 digits). If you only want significant digits to the right, you’d have to do something funny like

    >>> x = 1.23456789
    >>> # length of the number before the decimal
    >>> left_sigdigs = len(str(x).partition(".")[0])
    >>> format_string = "{0:0." + str(4 + left_sigdigs) + "g}"
    >>> print format_string.format(x)
    1.2346
    

    Note that it still rounds, unlike your example.

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