Convert floating point number to a certain precision, and then copy to string


I have a floating point number, say 135.12345678910. I want to concatenate that value to a string, but only want 135.123456789. With print, I can easily do this by doing something like:

print "%.9f" % numvar

with numvar being my original number. Is there an easy way to do this?

Asked By: pauliwago



It’s not print that does the formatting, It’s a property of strings, so you can just use

newstring = "%.9f" % numvar
Answered By: John La Rooy

Using round:

>>> numvar = 135.12345678910
>>> str(round(numvar, 9))
Answered By: shantanoo

With Python < 3 (e.g. 2.6 [see comments] or 2.7), there are two ways to do so.

# Option one
older_method_string = "%.9f" % numvar

# Option two
newer_method_string = "{:.9f}".format(numvar)

But note that for Python versions above 3 (e.g. 3.2 or 3.3), option two is preferred.

For more information on option two, I suggest this link on string formatting from the Python documentation.

And for more information on option one, this link will suffice and has info on the various flags.

Python 3.6 (officially released in December of 2016), added the f string literal, see more information here, which extends the str.format method (use of curly braces such that f"{numvar:.9f}" solves the original problem), that is,

# Option 3 (versions 3.6 and higher)
newest_method_string = f"{numvar:.9f}"

solves the problem. Check out @Or-Duan’s answer for more info, but this method is fast.

Answered By: jyalim

In case the precision is not known until runtime, this other formatting option is useful:

>>> n = 9
>>> '%.*f' % (n, numvar)
Answered By: Yu Hao

The str function has a bug. Please try the following. You will see ‘0,196553’ but the right output is ‘0,196554’. Because the str function’s default value is ROUND_HALF_UP.

>>> value=0.196553500000 
>>> str("%f" % value).replace(".", ",")
Answered By: ethemsulan

Python 3.6

Just to make it clear, you can use f-string formatting. This has almost the same syntax as the format method, but make it a bit nicer.



More reading about the new f string:

Here is a diagram of the execution times of the various tested methods (from last link above):

execution times

Answered By: Or Duan

To set precision with 9 digits, get:

print "%.9f" % numvar

Return precision with 2 digits:

print "%.2f" % numvar 

Return precision with 2 digits and float converted value:

numvar = 4.2345
print float("%.2f" % numvar) 
Answered By: Tejas Tank

It works as long as the number of decimals points are in range; After that it depends on the hardware. Beyond 14th decimal place, I could not get it to match. Values from latitude from a digital elevation file

lat1 = -81.0016666666670072 
lat2 = -81.0016666666670062
assert lat1 == lat2 # no asserion error :(

# try with Decimal
from decimal import *
getcontext().prec = 16
assert Decimal(lat1) == Decimal(lat2) # no asserion error :(

# Lets see string representation
print(f"{lat1:.16f}", f"{lat2:.16f}")
# -81.0016666666670062 -81.0016666666670062 :( Perils of Float

# Lets see how it is store in hardware
print(lat1.hex(), lat2.hex())
# -0x1.4401b4e81b500p+6 -0x1.4401b4e81b500p+6

# 14-th position it is able to identify 
lat1 = -81.0016666666670162 
lat2 = -81.0016666666670062
#-0x1.4401b4e81b501p+6 -0x1.4401b4e81b500p+6
print(lat1.hex(), lat2.hex())
assert lat1 == lat2 # assertion error :)
Answered By: Alex Punnen
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