Lambdas syntax error with if

Question

Let's take as an example with this simple dataset:

example_set  = pd.DataFrame(data = {"dirr":[1,0,-1,-1,1,-1,0], 
                                    "value": [125,130,80,8,150,251,18], 
                                    "result":[np.NaN for _ in range(7)]})

The following line returns error:invalid syntax

example_set["result"].apply(lambda x : example_set["value"]if x["dirr"]==1)

Can anyone tell me what I am doing wrong? Please do not come up with solution of how to do it without lambda: This is only a super simplified example that I made up to isolate my problem.

@EdChum Please read the question fully: Please do not come up with solution of how to do it without lambdas: This is only a super simplified example that I made up to isolate my problem. — jim jarnac
– jim jarnac, Commented Jan 2, 2017 at 16:04
.apply(lambda x : example_set["value"] if x["dirr"]==1 else x)? — Mohammad Yusuf
– Mohammad Yusuf, Commented Jan 2, 2017 at 16:06
@MYGz this returns another error: TypeError: 'float' object is not subscriptable — jim jarnac
– jim jarnac, Commented Jan 2, 2017 at 16:07

EdChum · Accepted Answer · 2017-01-02 16:15:42Z

8

You need to include the else return value in the lambda statement:

In [7]:
example_set['result'] = example_set.apply(lambda x: x['value'] if x['dirr'] == 1 else x['result'], axis = 1)
example_set

Out[7]:
   dirr  result  value
0     1   125.0    125
1     0     NaN    130
2    -1     NaN     80
3    -1     NaN      8
4     1   150.0    150
5    -1     NaN    251
6     0     NaN     18

Your attempt:

example_set["result"].apply(lambda x : example_set["value"]if x["dirr"]==1)

lacked the else return value in essence, also you needed x['value'] not the whole df, using the whole df yields a strange and undesirable return result even if the statement was corrected:

In [14]:
example_set.apply(lambda x: example_set['value'] if x['dirr'] == 1 else example_set['result'], axis = 1)

Out[14]:
       0      1     2    3      4      5     6
0  125.0  130.0  80.0  8.0  150.0  251.0  18.0
1    NaN    NaN   NaN  NaN    NaN    NaN   NaN
2    NaN    NaN   NaN  NaN    NaN    NaN   NaN
3    NaN    NaN   NaN  NaN    NaN    NaN   NaN
4  125.0  130.0  80.0  8.0  150.0  251.0  18.0
5    NaN    NaN   NaN  NaN    NaN    NaN   NaN
6    NaN    NaN   NaN  NaN    NaN    NaN   NaN

edited Jan 2, 2017 at 16:15

answered Jan 2, 2017 at 16:11

EdChum

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2 Comments

Mohammad Yusuf Over a year ago

Also, you can't access corresponding values of other columns if you .apply() on a single column. True?

EdChum Over a year ago

@MYGz yes that's correct, by definition it would call the Series version which has no axis param, only the DataFrame.apply version has axis param

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