Python print float without rounding

How can I take a float variable, and control how far out the float goes without round()? For example.

w = float(1.678)

I want to take x and make the following variables out of it.

x = 1.67
y = 1.6
z = 1

If I use the respective round methods:

x = round(w, 2) # With round I get 1.68 
y = round(y, 1) # With round I get 1.7
z = round(z, 0) # With round I get 2.0

It's going to round and alter the numbers to the point where there no use to me. I understand this is the point of round and its working properly. How would I go about getting the information that I need in the x,y,z variables and still be able to use them in other equations in a float format?

Mike Laren

7,84817 gold badges50 silver badges69 bronze badges

asked Mar 25, 2015 at 2:30

2

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(7)]
[1.0, 1.9, 1.92, 1.923, 1.9233, 1.92332, 1.923328]

answered Mar 25, 2015 at 14:00

4

A super simple solution is to use strings

x = float (str (w)[:-1])
y = float (str (w)[:-2])
z = float (str (w)[:-3])

Any of the floating point library solutions would require you dodge some rounding, and using floor/powers of 10 to pick out the decimals can get a little hairy by comparison to the above.

answered Mar 25, 2015 at 3:01

WakkaDojoWakkaDojo

4133 silver badges7 bronze badges

2

Integers are faster to manipulate than floats/doubles which are faster than strings. In this case, I tried to get time with both approach :

  timeit.timeit(stmt = "float(str(math.pi)[:12])", setup = "import math", number = 1000000)

~1.1929605630000424

for :

timeit.timeit(stmt = "math.floor(math.pi * 10 ** 10) / 10 ** 10", setup = "import math", number = 1000000)

~0.3455968870000561

So it's safe to use math.floor rather than string operation on it.

answered Feb 6, 2020 at 21:15

RajRaj

651 silver badge7 bronze badges

1

If you just need to control the precision in format

pi = 3.14159265
format(pi, '.3f') #print 3.142 # 3 precision after the decimal point
format(pi, '.1f') #print 3.1
format(pi, '.10f') #print 3.1415926500, more precision than the original

If you need to control the precision in floating point arithmetic

import decimal
decimal.getcontext().prec=4 #4 precision in total
pi = decimal.Decimal(3.14159265)
pi**2 #print Decimal('9.870') whereas '3.142 squared' would be off

--edit--

Without "rounding", thus truncating the number

import decimal
from decimal import ROUND_DOWN
decimal.getcontext().prec=4
pi*1 #print Decimal('3.142')

decimal.getcontext().rounding = ROUND_DOWN
pi*1 #print Decimal('3.141')

answered Mar 25, 2015 at 3:01

SYKSYK

6196 silver badges17 bronze badges

4

also this:

>>> f = 1.678
>>> n = 2
>>> int(f * 10 ** n) / 10 ** n
1.67

answered Jun 14 at 4:09

gtleegtlee

561 silver badge3 bronze badges

I think the easiest answer is :

from math import trunc

w = 1.678
x = trunc(w * 100) / 100
y = trunc(w * 10) / 10
z = trunc(w)

answered Jun 16 at 12:20

1

Easiest way to get integer:

series_col.round(2).apply(lambda x: float(str(x).split(".",1)[0]))

answered Jun 28 at 9:46

1

Not the answer you're looking for? Browse other questions tagged python python-2.7 or ask your own question.

How do you drop decimal places without rounding in Python?

How do I stop Python from rounding?

How do you print a float with 2 decimals in Python?

How do I restrict decimal places in Python?