Using list comprehension within lambda function?

Question

I have a dataframe with a list of n-gram tuples. I want to convert the list of tuples to list of strings. To do this, I am trying to use list comprehension within my lambda function. However, I keep getting an error saying that my list isn't defined.

id     n_grams
 1     [(thanks), (thanks, past), (thanks, past, blue)]
 2     [(support), (support, arm), (support, arm, brace), (support, arm, brace, left)]
 3     [(blue), (blue, sky), (blue, sky, rain)]
 4     [(breaking), (breaking, news), (breaking, news, fire), (breaking, news, fire, aparment)]

I am trying to get:

id     n_grams
 1     ["thanks", "thanks past", "thanks past blue"]
 2     ["support", "support arm", "support arm brace", "support arm brace left"]
 3     ["blue", "blue sky", "blue sky rain"]
 4     ["breaking", "breaking news", "breaking news fire", "breaking news fire apartment"]

I have tried:

data['n_grams'] = data.n_grams.apply(lambda row: " ".join(x) for x in row)

But I keep getting the error:

NameError: name 'row' is not defined

You mean .apply(lambda row: " ".join(x for x in row)) note, that is a generator expression, but it is totally pointless, it doesn't do anything. So just .apply(" ".join) would work — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Oct 28, 2021 at 14:45

mjoy · Accepted Answer · 2021-10-28 14:45:39Z

1

Thanks to the comment by rdas, the solution was simply using hard brackets:

data['n_grams'] = data.n_grams.apply(lambda row: [" ".join(x) for x in row])

answered Oct 28, 2021 at 14:45

mjoy

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atleastthreecharacters · Accepted Answer · 2021-10-28 14:54:56Z

0

This should work :

df.n_grams.apply(lambda row: [" ".join(x) for x in row])

answered Oct 28, 2021 at 14:54

atleastthreecharacters

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