Harvey
Harvey

Reputation: 637

Pandas logical operation

I'm having an issue with pandas 19.2 giving me the result I expect. The columns a-g have ['yes','no','', NaN]. If any of these columns have 'yes' I want the row returned (there are other columns not shown). Here is my code.

xdf2  =  xdf[((xdf['a'] == 'yes').all() or 
                    (xdf['b'] == 'yes').all() or 
                    (xdf['c'] == 'yes').all() or 
                    (xdf['d'] == 'yes' ).all() or
                    (xdf['e'] == 'yes').all() or 
                    (xdf['f'] == 'yes').all() or  
                    (xdf['g'] =='yes').all()) ]

This gives me the following error:

   2134                 return self._engine.get_loc(key)
   2135             except KeyError:
-> 2136                 return self._engine.get_loc(self._maybe_cast_indexer(key))
   2137 
   2138         indexer = self.get_indexer([key], method=method, tolerance=tolerance)

pandas\index.pyx in pandas.index.IndexEngine.get_loc (pandas\index.c:4433)()

pandas\index.pyx in pandas.index.IndexEngine.get_loc (pandas\index.c:4279)()

pandas\src\hashtable_class_helper.pxi in pandas.hashtable.PyObjectHashTable.get_item (pandas\hashtable.c:13742)()

pandas\src\hashtable_class_helper.pxi in pandas.hashtable.PyObjectHashTable.get_item (pandas\hashtable.c:13696)()

KeyError: False

Without the '.all' I get

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

This seems like a simple and common code snippet, but I haven't found a good example. What am I missing?

Upvotes: 1

Views: 241

Answers (2)

ImportanceOfBeingErnest
ImportanceOfBeingErnest

Reputation: 339795

The following should work:

import pandas as pd

a = [["yes", "no", "yes", "yes"],
     ["yes", "yes", "no", "yes"],
     ["yes", "no", "yes", "yes"]]
xdf = pd.DataFrame(a, columns=["a", "b", "c", "d"])     
print xdf

boollist = [ (xdf[col] == "yes").all() for col in xdf.columns ]
xdf2  =  xdf[xdf.columns[boollist] ]
print xdf2

Upvotes: 0

piRSquared
piRSquared

Reputation: 294586

Try

xdf[xdf[list('abcdefg')].eq('yes').any(1)]

Upvotes: 5

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