Python lambda function syntax to transform a pandas groupby dataframe

问题

This should be a very simple question to answer. I have two lines of code. The first one works. The second gives the following error:

SyntaxError: invalid syntax

Here are the two lines of code. The first line (which works fine) counts the rows where off0_on1 == 1. The second one trys to count the rows where off0_on1 == 0.

a1['on1'] = a1.groupby('del_month')['off0_on1'].transform(sum)
a1['off0'] = a1.groupby('del_month')['off0_on1'].transform(lambda x: 1 if x == 0)

Here is the pandas dataframe:

a1 = pd.DataFrame({'del_month':[1,1,1,1,2,2,2,2], 'off0_on1':[0,0,1,1,0,1,1,1]})

Any suggestions to revise the second line of code above?

Edit: Two of the answers have suggested using a map function, which produces the following output. The "on1" column is correct for my purposes; the "off0" column is not correct. For the first "del_month", the "off0" column should have the same results as the "on1" column. For the second "del_month", the "off0" column should be all ones (i.e. 1, 1, 1, 1).

Here's what happens when I use the following map function (see image below):

a1['off0'] = a1.groupby('del_month')['off0_on1'].transform(lambda series: map(lambda x: 1 if x == 0 else 0, series))

Edit 2 Not sure if this clarifies things, but ultimately I want pandas to do what the following SQL code does so easily:

select
    del_month
    , sum(case when off0_on1 = 1 then 1 else 0 end) as on1
    , sum(case when off0_on1 = 0 then 1 else 0 end) as off0
from a1
group by del_month
order by del_month

Edit 3 This new question contains the answer I need. Thanks everyone!

回答1:

When you define the if statement (shorthand way) you need to define the else as well.

lambda x: 1 if x == 0 else 0 # For example.

EDIT:

In your first lambda it's a Series Actually, so you need to loop it (with map for example)

a1['off0'] = a1.groupby('del_month')['off0_on1'].transform(lambda x: map(lambda value: 1 if value == 0 else 0, x))

回答2:

Your code: lambda x: 1 if x == 0 is not a valid expression. Try something like lambda x: 1 if x == 0 else 0

See below:

>>> example = lambda x: 1 if x == 0 else 0
>>> example(0)
1
>>> example(1)
0

回答3:

The previous answers quickly fixed my lambda function error by adding "else 0" at the end. My ultimate question was answered here with the following line of code:

a1['off0'] = a1.groupby('del_month')['off0_on1'].transform(lambda x: sum(x==0))

来源：https://stackoverflow.com/questions/53105016/python-lambda-function-syntax-to-transform-a-pandas-groupby-dataframe

标签

python

pandas

lambda

pandas-groupby