In Python, how can one print a number that might be an integer or real type, when the latter case would require me to limit my printout to a certain amount of digits?
Long story short, say we have the following example:
print("{0:.3f}".format(num)) # I cannot do print("{}".format(num))
# because I don't want all the decimals
Is there a "Pythy" way to ensure e.g. in case num == 1 that I print 1 instead of 1.000 (I mean other than cluttering my code with if statements)
With Python 3*, you can just use round() because in addition to rounding floats, when applied to an integer it will always return an int:
>>> num = 1.2345
>>> round(num,3)
1.234
>>> num = 1
>>> round(num,3)
1
This behavior is documented in help(float.__round__):
Help on method_descriptor:
__round__(...)
Return the Integral closest to x, rounding half toward even.
When an argument is passed, work like built-in round(x, ndigits).
And help(int.__round__):
Help on method_descriptor:
__round__(...)
Rounding an Integral returns itself.
Rounding with an ndigits argument also returns an integer.
* With Python 2, round() always returns a float.
If you need to maintain a fixed-width for float values, you could use the printf-style formatting, like this:
>>> num = 1
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1
>>> num = 1.2345
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.234
>>> num = 1.2
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.200
If you use a fix number of floating point, you could just use a replace to remove the extra 0. For instance this would do the trick:
print("{:.3f}".format(1).replace(".000", ""))
For fix number of decimal point:
>>> num = 0.2
>>> print('%.04*f' % num)
0.2000
>>> num = 3.102
>>> print('%.02*f' % num)
3.10
I am trying to do something that will convert any base to any other base without using decimal converting.
For example program will ask from user:
"Please put number which you would like to convert: 101"
"Please write from which base you want to convert: 2 " #base(2)
"Please write to which base you want to convert: 6" #base(6)
and will return value which converted from base(2) to base(6)
"Here is the result: 5"
You don't convert integers from one base to another. The integer doesn't change. You render an integer into a representation.
A normal (conventional) number is not, technically, a "decimal" number. It is a decimal representation of a number. Likewise for binary, octal, hexadecimal, etc.
For converting string representations of integers into any base between 2 and 36 you can simply use the int() built-in function with its extra (optional) argument. Python assumes a reasonably intuitive sequence of the letters 'a' through 'z' to represent "digits" greater than 9.
For example: int('101101',2) -> 45 and int('101101',3) -> 280 and so on.
For rendering an integer into binary, octal or hexadecimal one can simply use Python's built-in string formatting (though binary and perhaps octal were added sometime after version 2.2 of CPython ... and I don't remember the details).
It is, perhaps, not coincidental that Python's standard library definition of string.literal is returned in precisely the correct sequence to be used for any base up to 36 ... and one could, conceivably, use it to render integers into bases up through 99 ... though the notion of number representations where 'a' and 'A' are distinct digits along with spaces, newlines, etc. is a bit of a mind bender.
I have a naive approach, converting the number to base 10, our checkpoint, and then from ten.
def base(k,b,n):
def to_ten(k,b):
index = [
'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'
]
k = str(k)[::-1]
r = 0
for i in range(len(k)):
r += (b**i)*int(index.index(k[i]))
return r
def from_base(k,b):
r = ""
while k>0:
d = int(k%b)
if d<10:
r += str(d)
else:
r += chr(ord('A')+d-10)
k //= b
r = r[::-1]
return r
return from_base(to_ten(k,b),n)
k = input('Insert the number you\'d like to convert: ')
b = int(input('Insert the base you\'d like to convert ' + k + ' from: '))
n = int(input('Insert the base you\'d like to convert to: '))
print('Your answer is: ' + str(base(k,b,n)))
Please tell me if the code does not work.
I want to remove digits from a float to have a fixed number of digits after the dot, like:
1.923328437452 → 1.923
I need to output as a string to another function, not print.
Also I want to ignore the lost digits, not round them.
round(1.923328437452, 3)
See Python's documentation on the standard types. You'll need to scroll down a bit to get to the round function. Essentially the second number says how many decimal places to round it to.
First, the function, for those who just want some copy-and-paste code:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '{}'.format(f)
if 'e' in s or 'E' in s:
return '{0:.{1}f}'.format(f, n)
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
This is valid in Python 2.7 and 3.1+. For older versions, it's not possible to get the same "intelligent rounding" effect (at least, not without a lot of complicated code), but rounding to 12 decimal places before truncation will work much of the time:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '%.12f' % f
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
Explanation
The core of the underlying method is to convert the value to a string at full precision and then just chop off everything beyond the desired number of characters. The latter step is easy; it can be done either with string manipulation
i, p, d = s.partition('.')
'.'.join([i, (d+'0'*n)[:n]])
or the decimal module
str(Decimal(s).quantize(Decimal((0, (1,), -n)), rounding=ROUND_DOWN))
The first step, converting to a string, is quite difficult because there are some pairs of floating point literals (i.e. what you write in the source code) which both produce the same binary representation and yet should be truncated differently. For example, consider 0.3 and 0.29999999999999998. If you write 0.3 in a Python program, the compiler encodes it using the IEEE floating-point format into the sequence of bits (assuming a 64-bit float)
0011111111010011001100110011001100110011001100110011001100110011
This is the closest value to 0.3 that can accurately be represented as an IEEE float. But if you write 0.29999999999999998 in a Python program, the compiler translates it into exactly the same value. In one case, you meant it to be truncated (to one digit) as 0.3, whereas in the other case you meant it to be truncated as 0.2, but Python can only give one answer. This is a fundamental limitation of Python, or indeed any programming language without lazy evaluation. The truncation function only has access to the binary value stored in the computer's memory, not the string you actually typed into the source code.1
If you decode the sequence of bits back into a decimal number, again using the IEEE 64-bit floating-point format, you get
0.2999999999999999888977697537484345957637...
so a naive implementation would come up with 0.2 even though that's probably not what you want. For more on floating-point representation error, see the Python tutorial.
It's very rare to be working with a floating-point value that is so close to a round number and yet is intentionally not equal to that round number. So when truncating, it probably makes sense to choose the "nicest" decimal representation out of all that could correspond to the value in memory. Python 2.7 and up (but not 3.0) includes a sophisticated algorithm to do just that, which we can access through the default string formatting operation.
'{}'.format(f)
The only caveat is that this acts like a g format specification, in the sense that it uses exponential notation (1.23e+4) if the number is large or small enough. So the method has to catch this case and handle it differently. There are a few cases where using an f format specification instead causes a problem, such as trying to truncate 3e-10 to 28 digits of precision (it produces 0.0000000002999999999999999980), and I'm not yet sure how best to handle those.
If you actually are working with floats that are very close to round numbers but intentionally not equal to them (like 0.29999999999999998 or 99.959999999999994), this will produce some false positives, i.e. it'll round numbers that you didn't want rounded. In that case the solution is to specify a fixed precision.
'{0:.{1}f}'.format(f, sys.float_info.dig + n + 2)
The number of digits of precision to use here doesn't really matter, it only needs to be large enough to ensure that any rounding performed in the string conversion doesn't "bump up" the value to its nice decimal representation. I think sys.float_info.dig + n + 2 may be enough in all cases, but if not that 2 might have to be increased, and it doesn't hurt to do so.
In earlier versions of Python (up to 2.6, or 3.0), the floating point number formatting was a lot more crude, and would regularly produce things like
>>> 1.1
1.1000000000000001
If this is your situation, if you do want to use "nice" decimal representations for truncation, all you can do (as far as I know) is pick some number of digits, less than the full precision representable by a float, and round the number to that many digits before truncating it. A typical choice is 12,
'%.12f' % f
but you can adjust this to suit the numbers you're using.
1Well... I lied. Technically, you can instruct Python to re-parse its own source code and extract the part corresponding to the first argument you pass to the truncation function. If that argument is a floating-point literal, you can just cut it off a certain number of places after the decimal point and return that. However this strategy doesn't work if the argument is a variable, which makes it fairly useless. The following is presented for entertainment value only:
def trunc_introspect(f, n):
'''Truncates/pads the float f to n decimal places by looking at the caller's source code'''
current_frame = None
caller_frame = None
s = inspect.stack()
try:
current_frame = s[0]
caller_frame = s[1]
gen = tokenize.tokenize(io.BytesIO(caller_frame[4][caller_frame[5]].encode('utf-8')).readline)
for token_type, token_string, _, _, _ in gen:
if token_type == tokenize.NAME and token_string == current_frame[3]:
next(gen) # left parenthesis
token_type, token_string, _, _, _ = next(gen) # float literal
if token_type == tokenize.NUMBER:
try:
cut_point = token_string.index('.') + n + 1
except ValueError: # no decimal in string
return token_string + '.' + '0' * n
else:
if len(token_string) < cut_point:
token_string += '0' * (cut_point - len(token_string))
return token_string[:cut_point]
else:
raise ValueError('Unable to find floating-point literal (this probably means you called {} with a variable)'.format(current_frame[3]))
break
finally:
del s, current_frame, caller_frame
Generalizing this to handle the case where you pass in a variable seems like a lost cause, since you'd have to trace backwards through the program's execution until you find the floating-point literal which gave the variable its value. If there even is one. Most variables will be initialized from user input or mathematical expressions, in which case the binary representation is all there is.
The result of round is a float, so watch out (example is from Python 2.6):
>>> round(1.923328437452, 3)
1.923
>>> round(1.23456, 3)
1.2350000000000001
You will be better off when using a formatted string:
>>> "%.3f" % 1.923328437452
'1.923'
>>> "%.3f" % 1.23456
'1.235'
n = 1.923328437452
str(n)[:4]
At my Python 2.7 prompt:
>>> int(1.923328437452 * 1000)/1000.0
1.923
The truely pythonic way of doing it is
from decimal import *
with localcontext() as ctx:
ctx.rounding = ROUND_DOWN
print Decimal('1.923328437452').quantize(Decimal('0.001'))
or shorter:
from decimal import Decimal as D, ROUND_DOWN
D('1.923328437452').quantize(D('0.001'), rounding=ROUND_DOWN)
Update
Usually the problem is not in truncating floats itself, but in the improper usage of float numbers before rounding.
For example: int(0.7*3*100)/100 == 2.09.
If you are forced to use floats (say, you're accelerating your code with numba), it's better to use cents as "internal representation" of prices: (70*3 == 210) and multiply/divide the inputs/outputs.
Simple python script -
n = 1.923328437452
n = float(int(n * 1000))
n /=1000
def trunc(num, digits):
sp = str(num).split('.')
return '.'.join([sp[0], sp[1][:digits]])
This should work. It should give you the truncation you are looking for.
So many of the answers given for this question are just completely wrong. They either round up floats (rather than truncate) or do not work for all cases.
This is the top Google result when I search for 'Python truncate float', a concept which is really straightforward, and which deserves better answers. I agree with Hatchkins that using the decimal module is the pythonic way of doing this, so I give here a function which I think answers the question correctly, and which works as expected for all cases.
As a side-note, fractional values, in general, cannot be represented exactly by binary floating point variables (see here for a discussion of this), which is why my function returns a string.
from decimal import Decimal, localcontext, ROUND_DOWN
def truncate(number, places):
if not isinstance(places, int):
raise ValueError("Decimal places must be an integer.")
if places < 1:
raise ValueError("Decimal places must be at least 1.")
# If you want to truncate to 0 decimal places, just do int(number).
with localcontext() as context:
context.rounding = ROUND_DOWN
exponent = Decimal(str(10 ** - places))
return Decimal(str(number)).quantize(exponent).to_eng_string()
>>> from math import floor
>>> floor((1.23658945) * 10**4) / 10**4
1.2365
# divide and multiply by 10**number of desired digits
If you fancy some mathemagic, this works for +ve numbers:
>>> v = 1.923328437452
>>> v - v % 1e-3
1.923
I did something like this:
from math import trunc
def truncate(number, decimals=0):
if decimals < 0:
raise ValueError('truncate received an invalid value of decimals ({})'.format(decimals))
elif decimals == 0:
return trunc(number)
else:
factor = float(10**decimals)
return trunc(number*factor)/factor
You can do:
def truncate(f, n):
return math.floor(f * 10 ** n) / 10 ** n
testing:
>>> f=1.923328437452
>>> [truncate(f, n) for n in range(5)]
[1.0, 1.9, 1.92, 1.923, 1.9233]
Just wanted to mention that the old "make round() with floor()" trick of
round(f) = floor(f+0.5)
can be turned around to make floor() from round()
floor(f) = round(f-0.5)
Although both these rules break around negative numbers, so using it is less than ideal:
def trunc(f, n):
if f > 0:
return "%.*f" % (n, (f - 0.5*10**-n))
elif f == 0:
return "%.*f" % (n, f)
elif f < 0:
return "%.*f" % (n, (f + 0.5*10**-n))
def precision(value, precision):
"""
param: value: takes a float
param: precision: int, number of decimal places
returns a float
"""
x = 10.0**precision
num = int(value * x)/ x
return num
precision(1.923328437452, 3)
1.923
Short and easy variant
def truncate_float(value, digits_after_point=2):
pow_10 = 10 ** digits_after_point
return (float(int(value * pow_10))) / pow_10
>>> truncate_float(1.14333, 2)
>>> 1.14
>>> truncate_float(1.14777, 2)
>>> 1.14
>>> truncate_float(1.14777, 4)
>>> 1.1477
When using a pandas df this worked for me
import math
def truncate(number, digits) -> float:
stepper = 10.0 ** digits
return math.trunc(stepper * number) / stepper
df['trunc'] = df['float_val'].apply(lambda x: truncate(x,1))
df['trunc']=df['trunc'].map('{:.1f}'.format)
int(16.5);
this will give an integer value of 16, i.e. trunc, won't be able to specify decimals, but guess you can do that by
import math;
def trunc(invalue, digits):
return int(invalue*math.pow(10,digits))/math.pow(10,digits);
Here is an easy way:
def truncate(num, res=3):
return (floor(num*pow(10, res)+0.5))/pow(10, res)
for num = 1.923328437452, this outputs 1.923
def trunc(f,n):
return ('%.16f' % f)[:(n-16)]
A general and simple function to use:
def truncate_float(number, length):
"""Truncate float numbers, up to the number specified
in length that must be an integer"""
number = number * pow(10, length)
number = int(number)
number = float(number)
number /= pow(10, length)
return number
There is an easy workaround in python 3. Where to cut I defined with an help variable decPlace to make it easy to adapt.
f = 1.12345
decPlace= 4
f_cut = int(f * 10**decPlace) /10**decPlace
Output:
f = 1.1234
Hope it helps.
Most answers are way too complicated in my opinion, how about this?
digits = 2 # Specify how many digits you want
fnum = '122.485221'
truncated_float = float(fnum[:fnum.find('.') + digits + 1])
>>> 122.48
Simply scanning for the index of '.' and truncate as desired (no rounding).
Convert string to float as final step.
Or in your case if you get a float as input and want a string as output:
fnum = str(122.485221) # convert float to string first
truncated_float = fnum[:fnum.find('.') + digits + 1] # string output
I think a better version would be just to find the index of decimal point . and then to take the string slice accordingly:
def truncate(number, n_digits:int=1)->float:
'''
:param number: real number ℝ
:param n_digits: Maximum number of digits after the decimal point after truncation
:return: truncated floating point number with at least one digit after decimal point
'''
decimalIndex = str(number).find('.')
if decimalIndex == -1:
return float(number)
else:
return float(str(number)[:decimalIndex+n_digits+1])
int(1.923328437452 * 1000) / 1000
>>> 1.923
int(1.9239 * 1000) / 1000
>>> 1.923
By multiplying the number by 1000 (10 ^ 3 for 3 digits) we shift the decimal point 3 places to the right and get 1923.3284374520001. When we convert that to an int the fractional part 3284374520001 will be discarded. Then we undo the shifting of the decimal point again by dividing by 1000 which returns 1.923.
use numpy.round
import numpy as np
precision = 3
floats = [1.123123123, 2.321321321321]
new_float = np.round(floats, precision)
Something simple enough to fit in a list-comprehension, with no libraries or other external dependencies. For Python >=3.6, it's very simple to write with f-strings.
The idea is to let the string-conversion do the rounding to one more place than you need and then chop off the last digit.
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1] for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.800', '0.888', '1.125', '1.250', '1.500']
Of course, there is rounding happening here (namely for the fourth digit), but rounding at some point is unvoidable. In case the transition between truncation and rounding is relevant, here's a slightly better example:
>>> nacc = 6 # desired accuracy (maximum 15!)
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nacc}f}'[:-(nacc-nout)] for x in [2.9999, 2.99999, 2.999999, 2.9999999]]
>>> ['2.999', '2.999', '2.999', '3.000']
Bonus: removing zeros on the right
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1].rstrip('0') for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.8', '0.888', '1.125', '1.25', '1.5']
The core idea given here seems to me to be the best approach for this problem.
Unfortunately, it has received less votes while the later answer that has more votes is not complete (as observed in the comments). Hopefully, the implementation below provides a short and complete solution for truncation.
def trunc(num, digits):
l = str(float(num)).split('.')
digits = min(len(l[1]), digits)
return l[0] + '.' + l[1][:digits]
which should take care of all corner cases found here and here.
Am also a python newbie and after making use of some bits and pieces here, I offer my two cents
print str(int(time.time()))+str(datetime.now().microsecond)[:3]
str(int(time.time())) will take the time epoch as int and convert it to string and join with...
str(datetime.now().microsecond)[:3] which returns the microseconds only, convert to string and truncate to first 3 chars
# value value to be truncated
# n number of values after decimal
value = 0.999782
n = 3
float(int(value*1en))*1e-n
I'm working on a script that needs to output numbers in a precision determined by length and not just decimal places. Let's say I want a max of 7 decimals but whole numbers above 7 digits are ok although they can't have decimals if the total amount of digits is 7 or higher.
So the rules are:
- If a float has more than 7 digits of whole numbers, it loses it's decimals
- If a float has less than 7 digits of whole numbers, it gets a length of 7 total digits including decimals
- If a floats has no digits of whole numbers, it keeps 7 decimals
For example:
a = 10000.02
b = 1.000002
c = 100000000000
are all correct.
How do I handle numbers (all float) in this format?
For example:
d = 892432589.54680
e = 382.39810758264057251
f = 452545871.1643548
g = 10254.968743541742
h = 165783438364325.126
I was thinking about something along the lines of:
length_of_number = len(str(x))
if length_of_number > 7:
x.round(0)
else:
x.round(7 - length_of_number)
but then I get in trouble with situations like
x= 0.5313231568943218748
because the whole float is longer than 7 digits.
What to do?
Edit for asked examples:
12345678 is ok
1234567.1 is ok (but 1234567.12 not)
1.12345678 would become 1.123456
I don't really get what you want, but perhaps this can help you?
round(num_here, 7)
From the documentation:
Return number rounded to ndigits precision after the decimal point. If ndigits is omitted or is None, it returns the nearest integer to its input.
For the built-in types supporting round() , values are rounded to the closest multiple of 10 to the power minus ndigits; if two multiples are equally close, rounding is done toward the even choice (so, for example, both round(0.5) and round(-0.5) are 0, and round(1.5) is 2). Any integer value is valid for ndigits (positive, zero, or negative). The return value is an integer if called with one argument, otherwise of the same type as number.
For a general Python object number, round(number, ndigits) delegates to number.__round__(ndigits).
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.
I think what you are asking can be solved in how you determine your length and also by finding if there is a decimal in your string or not. For instance if you said
>>>a = 1325.6326
>>>len(str(a).split(".")[0])
4
This will give you the length of all your numbers to the left of the decimal place.
>>>b = '.' in str(a)
>>>print(b)
True
And this will tell you if there is a decimal place in your number.
def truncate(n):
s = str(n)
is_decimal = '.' in s
if len(s) <= 7 + is_decimal:
return s
if not is_decimal:
return s
fore, aft = s.split('.', 1)
if len(fore) >= 7:
return fore
if n < 1:
return '{:.7f}'.format(n)
return f'{{:.{7-len(fore)}f}}'.format(n)
Here's one version, using string formatting to truncate the decimals. It turns out that string formatting actually rounds, instead of dropping characters, so string slicing may work better for you
def truncate(n):
s = str(n)
is_decimal = '.' in s
if len(s) <= 7 + is_decimal:
return s
if not is_decimal:
return s
fore, aft = s.split('.', 1)
if len(fore) >= 7:
return fore
if n < 1:
return '.' + aft[:7]
return '{}.{}'.format(fore, aft[:7-len(fore)])
I change your code slightly and turned it into a function to make it easier to test.
num = [892432589.54680,
382.39810758264057251,
452545871.1643548,
10254.968743541742,
165783438364325.126,
0.5313231568943218748]
def trunc_num(x):
num_int = int(x)
length_of_number = len(str(num_int))
if length_of_number >= 7:
return num_int
else:
return round(x, 7 - length_of_number)
for n in num:
print(trunc_num(n))
By changing the number into an int the decimals are dropped and we can check the length of the number with out the decimal in the way. If the int is greater than 7 digits we just return the int else we round the number as you were trying before.
Output
892432589
382.3981
452545871
10254.97
165783438364325
0.531323
If you need to account for negative numbers than you can just add lstrip after converting it to a string.
length_of_number = len(str(num_int).lstrip('-'))
Here's a super simple version using string slicing:
def seven(x):
ret = lambda x: float(x) if '.' in str(x) else int(x)
return ret(str(x)[:max(str(x).find('.'), 8 + ('-' in str(x)))].strip('.'))
Output:
<class 'int'> 892432590
<class 'int'> -892432590
<class 'float'> 382.3981
<class 'float'> -382.3981
<class 'int'> 452545871
<class 'int'> -452545871
<class 'float'> 10254.97
<class 'float'> -10254.97
<class 'int'> 165783438364325
<class 'int'> -165783438364325
<class 'float'> 0.531323
<class 'float'> -0.531323
An alternative using str.format():
def seven(x):
ret = lambda x: float(x) if '.' in x else int(x)
return ret('{:.{cutoff}f}'.format(x, cutoff=7 - min(str(x).find('.') - ('-' in str(x)), 7)))