This question already has answers here:
How to truncate float values?
(31 answers)
Closed 9 years ago.
I've found dozen of answers, but non of them is what I'm looking for, I don't want to round up or down, I know that I can round numbers as follow:
>>> print('%.3f' % 15.555555)
15.556
>>> round(15.555555, 3)
15.666
But I need to get 15.555. Should I use regex?
Cheeky solution:
numstring = str(15.555555)
num = float(numstring[:numstring.find('.')+4])
My solution involving int abuse. int rounds towards the nearest 0. I multiply it by 10**3 to affect rounding. After using int, I divide it by 10**3 to get actual results.
It's safer, as it does work with e notation.
int(15.55555 * 10**3) / 10.0**3
Related
This question already has answers here:
Limiting floats to two decimal points
(35 answers)
Closed last year.
I have two float values 1000.4 and 700.7
The result of 1000.4 - 700.7 returns me 299.69999999999993. I did my research,Decimal would be a good option to do the calculation here print(Decimal('1000.4') - Decimal('700.7')) and it returns 299.7
I have a question. what if I have a value 14.5 how can I print it as 14.50?
I tried getcontext().prec = 2 and it didn't help and would made the print(Decimal('1000.4') - Decimal('700.7')) returns 3.0E+2 which isn't what I want.
You can use f-strings to custom the leading zeros/limit the precision.
n=14.5
print(f"{n:.2f}")
Here, it would print only the first two decimals. (14.50).
This can be done with formatting for both floats and Decimal.
In [1]: print("{:.2f}".format(1.234))
1.23
In [2]: from decimal import Decimal
In [3]: print("{:.2f}".format(Decimal(1.234)))
1.23
This question already has answers here:
Python round up integer to next hundred
(10 answers)
Closed 2 years ago.
The built-in function round() will round a value down but I want to know how to round a value up.
i know that this is possible with math.ceil() but the thing is that round() has the keyword argument "ndigits" and math.ceil() doesn't. so for example:
>>> round(1024, ndigits=-3)
1000
but i want 1100.
Is there a solution for this?
import math
def round(number, n):
return math.ceil(number * math.pow(10, n+1))*(math.pow(10,-(n+1)))
print(round(1024, -3))
# 1100.0
A simple function like this would suffice, multiply by 10^(n+1), find the ceiling of that number, then multiply by 10^-n-1 (equivalent to dividing by 10^n+1).
This question already has answers here:
Rounding a number in Python but keeping ending zeros
(6 answers)
Closed 6 years ago.
So let's say I have this code:
num = 1.29283
round(num, 2)
That rounds to 1.29, but if I do this:
num = 1.30293
round(num, 2)
That rounds to 1.3. I want to know if there is a way to have it round to 1.30; I know it is the same number, but I need it to print 1.30.
You can use string formatting for this. A number in python does not have such a thing as trailing zeros. So your question only make sense for strings.
Example:
>>> num = 1.30293
>>> "{:.2f}".format(num)
'1.30'
The .2f says that this is a float (f) and that you want two digits after the point .2. Read more about string formatting here
This question already has answers here:
Negative integer division surprising result
(5 answers)
Closed 7 years ago.
How python calculate this division?
>>>-3/10
-1
Looks like python rounds the answer to the lower value.
>>> -3/4
-1
>>> -3/4.
-0.75
>>> -3/10.
-0.3
>>> -3/10
-1
This is just my guess.
Python 2, like many languages, uses integer division. Dividing two integers returns a integer (the nearest integer to the answer rounded down.)
To get a floating point result, you need to force one or more of the terms to be a float.
float(-3)/10
This question already has answers here:
How to display a float with two decimal places?
(13 answers)
Pad python floats
(4 answers)
Closed 8 years ago.
I'm sorry, I know this must be a duplicate, I can't find where else it's posted. Please feel free to link me to the original question and mark this as duplicate.
I would like to print a 3 digits of a number AFTER the decimal point in it.
For example:
number = 523.637382
I would like to print: 523.637
I have a feeling I can use something similar to this
print(str(number)[:7])
>>>523.637
However, this will not work if the number before the decimal is not 3 decimals.
Bonus points:
Would this be easy?
number = 500.220
#magic
>>>500.22
number = 500.2000003
#magic
>>>500.2
A (built-in) function that could do this is round:
>>> number = 523.637382
>>> rounded = round(number, 3) # 3 decimal places, for example
>>> rounded
523.637
This has already been answered for example here.
The good news, to answer the second part of your question, is that the round function automatically removes trailing zeroes. It's much harder to retain the zeros if you're defining a new variable: you need the decimal module; but it looks that that isn't necessary here.
>>> number = 523.60000001
>>> rounded = round(number, 3)
>>> rounded
523.6
print("%.3f" % number)
or, using the new-style formatting,
print("{0:.3f}".format(number))
If you're printing a str like above you can use string interpolation:
number = 33.33333
print("{0:.3f}".format(number))
#=> 33.333