You can use the normalize method to remove extra precision.
>>> print decimal.Decimal('5.500')
5.500
>>> print decimal.Decimal('5.500').normalize()
5.5
To avoid stripping zeros to the left of the decimal point, you could do this:
def normalize_fraction(d):
normalized = d.normalize()
sign, digits, exponent = normalized.as_tuple()
if exponent > 0:
return decimal.Decimal((sign, digits + (0,) * exponent, 0))
else:
return normalized
Or more compactly, using quantize as suggested by user7116:
def normalize_fraction(d):
normalized = d.normalize()
sign, digit, exponent = normalized.as_tuple()
return normalized if exponent <= 0 else normalized.quantize(1)
You could also use to_integral() as shown here but I think using as_tuple this way is more self-documenting.
I tested these both against a few cases; please leave a comment if you find something that doesn't work.
>>> normalize_fraction(decimal.Decimal('55.5'))
Decimal('55.5')
>>> normalize_fraction(decimal.Decimal('55.500'))
Decimal('55.5')
>>> normalize_fraction(decimal.Decimal('55500'))
Decimal('55500')
>>> normalize_fraction(decimal.Decimal('555E2'))
Decimal('55500')
You can use the normalize method to remove extra precision.
>>> print decimal.Decimal('5.500')
5.500
>>> print decimal.Decimal('5.500').normalize()
5.5
To avoid stripping zeros to the left of the decimal point, you could do this:
def normalize_fraction(d):
normalized = d.normalize()
sign, digits, exponent = normalized.as_tuple()
if exponent > 0:
return decimal.Decimal((sign, digits + (0,) * exponent, 0))
else:
return normalized
Or more compactly, using quantize as suggested by user7116:
def normalize_fraction(d):
normalized = d.normalize()
sign, digit, exponent = normalized.as_tuple()
return normalized if exponent <= 0 else normalized.quantize(1)
You could also use to_integral() as shown here but I think using as_tuple this way is more self-documenting.
I tested these both against a few cases; please leave a comment if you find something that doesn't work.
>>> normalize_fraction(decimal.Decimal('55.5'))
Decimal('55.5')
>>> normalize_fraction(decimal.Decimal('55.500'))
Decimal('55.5')
>>> normalize_fraction(decimal.Decimal('55500'))
Decimal('55500')
>>> normalize_fraction(decimal.Decimal('555E2'))
Decimal('55500')
Answer from the Decimal FAQ in the documentation:
>>> def remove_exponent(d):
... return d.quantize(Decimal(1)) if d == d.to_integral() else d.normalize()
>>> remove_exponent(Decimal('5.00'))
Decimal('5')
>>> remove_exponent(Decimal('5.500'))
Decimal('5.5')
>>> remove_exponent(Decimal('5E+3'))
Decimal('5000')
Hello everyone,
I am still new to python and learning.
So I practiced some exercises and made an app that calculates the percentage from the number the user enters.
My question use, how can I terminate the .0 part if the user enters an Int and keep the decimal part if they enter a float?
so for example, 5% of 100 is 5 ( Int)
and 5.1% of 100 is 5.1 (float)
I have a float formatted to 2 decimal places. I need to eliminate the 2nd decimal place if it's a "0" but still keep 2 decimal places open for when its 2 whole numbers.
number = float(25.20458)
print(format(number, ".2f"))
#Comes out as 25.20
#Need 25.2Windows 10 and Python 3.7
If want convert integers and floats numbers to strings with no trailing 0 use this with map or apply:
df = pd.DataFrame({'col1':[1.00, 1, 0.5, 1.50]})
df['new'] = df['col1'].map('{0:g}'.format)
#alternative solution
#df['new'] = df['col1'].apply('{0:g}'.format)
print (df)
col1 new
0 1.0 1
1 1.0 1
2 0.5 0.5
3 1.5 1.5
print (df['new'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
3 <class 'str'>
Name: new, dtype: object
I think something like this should work:
if val.is_integer() == True :
val = int(val)
elif val.is_float() == True :
val = Decimal(val).normalize()
Assuming that val is a float value inside the dataframe's column. You simply cast the value to be integer.
For float value instead you cut extra zeros.
You would need to reassign x to the value of x = int(x) or you could also use str.format if you just want the output formatted:
print "Het antwoord van de berekening is: {:.0f}.".format(x)
int and round will exhibit different behaviour, if you have anything >= 5 after the decimal point then int will floor but round will round up, if you want to actually use round you might want to combine the two:
In [7]: x = round(1.5)
In [8]: x
Out[8]: 2.0
In [9]: int(x)
Out[9]: 2
Or again combine with str.format:
In [10]: print "Het antwoord van de berekening is: {:.0f}".format(round(1.5))
Het antwoord van de berekening is: 2
The round() function cannot alter the x variable in place, as numbers are immutable. Instead, the rounded result is returned, which your code ignores.
Store the result back in x:
x = round(x)
This will give you a floating point number rounded to the nearest whole number.
Alternatively, use x = int(x), which gives you an integer number, but floors that number (removes the decimal portion regardless if it is closer to the next whole number or not).
Here's a function to format your numbers the way you want them:
def formatNumber(num):
if num % 1 == 0:
return int(num)
else:
return num
For example:
formatNumber(3.11111)
returns
3.11111
formatNumber(3.0)
returns
3
you can use string formatting
>>> "%g" % 1.1
'1.1'
>>> "%g" % 1.0
'1'
What about converting it to int?
>>>int(a)
100
Just for the sake of completeness, there are many many ways to remove the decimal part from a string representation of a decimal number, one that I can come up right now is:
s='100.0'
s=s[:s.index('.')]
s
>>>'100'
Perhaps there's another one more simple.
Hope this helps!
If you do not want to convert it to an int you can also split it.
>>> a = 100.25
>>> str(a).split('.')[0]
>>> '100' # result is now a string