Passing strings in a lambda function

Question

I would like to automate the following code line:

df_b = df_b.append([{'A':df_a['A'][1], 'B':df_a['B'][1], 'C':df_a['C'][1]}], ignore_index=True)

and I was considering using a lambda function to do so:

list = ['A', 'B', 'C']
func = lambda j: df_a[j][1]
df_b = df_b.append([{(func(list))}], ignore_index=True)

And I got the following error:

KeyError                                  Traceback (most recent call last)
~\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, tolerance)
   2896             try:
-> 2897                 return self._engine.get_loc(key)
   2898             except KeyError:

pandas\_libs\index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas\_libs\index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas\_libs\hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

pandas\_libs\hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

KeyError: 1

What would be your advice to run such a lambda function? Thanks

bjdev · Accepted Answer · 2022-10-21 08:38:51Z

1

I think what you want here is a dictionary comprehension:

lst = ['A', 'B', 'C']
func = lambda j: df_a[j][1]
to_append = {s:func(s) for s in lst}
df_b = df_b.append([to_append], ignore_index=True)

As a side note, you could equally define func with a def instead of lambda in this case.

answered Oct 21, 2022 at 8:38

bjdev

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1 Comment

manu190466 Over a year ago

Your answer is OK, but the lambda function is so trivial here that it would be clearer to directly implement its content in the dictionnary comprehension.

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