replace df index values with values from a list but ignore empty strings

Question

I want to replace DF1.index.values with DF2[some_column].values HOWEVER i only want to replace if DF2[some_column].value is not null or empty string.

DF1.index.values

Index(['a','b','c','d']), dtype='object')

DF2[some_column].values

['base','','','net 1']

Expected output

Index(['base','b','c','net 1']), dtype='object')

My attempt:

DF1.index = DF2[some_column].values

It's incorrect because it replaces everything and thats not what I want, I'm only interested in not null or empty values.

does `DF2[some_column].replace('', DF1.index)` work? – EdChum Jan 22 '16 at 13:05 — EdChum, Jan 22 '16 at 13:05
Thanks EdChum but it wouldnt do fulfill what I plan to do. – Boosted_d16 Jan 22 '16 at 16:47 — Boosted_d16, Jan 22 '16 at 16:47

score 1 · Accepted Answer · answered Jan 22 '16 at 13:35

You can use where to choose values by the condition from to sources:

DF1.index = DF2[some_column].where(DF2[some_column]!="", DF2.index)

So it will use DF2[some_column] values if DF2[some_column]!="" (in fact here we create a Series which is used as boolean index saying what data to choose for each raw), in the other case it will use DF2.index value.

As you don't want nan values also, you need pd.notnull functions. And to check both conditions use "logical and" function for pairs of values in two Series which is &.

So the final code is

DF1.index = DF2[some_column].where(
           (DF2[some_column]!="") & pd.notnull(DF2[some_column]), DF2.index)

replace df index values with values from a list but ignore empty strings

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