I am trying to divide the numbers in two columns, as a percent and catch any ZeroDivisionErrors. I have been trying to figure out the formatting works, and so far nothing has worked.
import pandas as pd
def percent_diff(col1, col2):
"""
This function will avoid any error caused by a divide by 0. The result will be 1,
representing that there is a 100 % difference
"""
try:
x = 1 - (col1 / col2)
x = "{:,.2%}".format(x)
return x
except ZeroDivisionError:
x = 'Error'
return x
data = {'a' : [1, 2, 3, 4, 5, 6, 7, 0, 9, 10],
'b' : [10, 9, 0, 7, 6, 5, 4, 3, 2, 1]}
df = pd.DataFrame(data)
df['c'] = percent_diff(df['a'], df['b'])
df.head(10)
I would like another column with a percent like 25.12%
, or Error
if there is a division error. 100.00% would also work in my instance.
CodePudding user response:
A straightforward solution using zip:
[a / b for (a, b) in zip(la, lb)]
You can replace the a / b
with a function call (percent_diff
) that handles your zero division case, like you have, but without needing to manage the list iteration.
That said, zip
will zip up two lists into a tuple that you can utilize:
>>> la = [1,2,3]
>>> lb = [2,2,2]
>>> [i for i in zip(la, lb)]
[(1, 2), (2, 2), (3, 2)]
>>> [a / b for (a, b) in zip(la, lb)]
[0.5, 1.0, 1.5]
A full solution looks like:
def perc(a, b):
try:
result = 1 - (a / b) # Note this inverts the percentage.
return "{:,.2%}".format(result)
except ZeroDivisionError:
return "Error"
data = {'a' : [1, 2, 3, 4, 5, 6, 7, 0, 9, 10],
'b' : [10, 9, 0, 7, 6, 5, 4, 3, 2, 1]}
data["c"] = [perc(a, b) for (a, b) in zip(data.get("a", []), data.get("b", []))]
Yielding the result:
>>> pprint(data)
{'a': [1, 2, 3, 4, 5, 6, 7, 0, 9, 10],
'b': [10, 9, 0, 7, 6, 5, 4, 3, 2, 1],
'c': ['90.00%',
'77.78%',
'Error',
'42.86%',
'16.67%',
'-20.00%',
'-75.00%',
'100.00%',
'-350.00%',
'-900.00%']}
CodePudding user response:
You're passing a pd.Series into the format, which isn't supported apparently.
This answer shows you can just use map()
df['c'] = (1-df['a']/df['b']).map('%{:,.2%}'.format)
CodePudding user response:
You could save yourself the exception management by computing your expected result directly:
return f"{1-(col1/col2):,.2%}" if col2 else "Error"
or (to abide by the function's comment)
return f"{1-(col1/col2):,.2%}" if col2 else "100%"