CODEWITHSUNDEEP
pandaspivot-table
william007
william007

Reputation: 18547

ValueError: Grouper for 'C' not 1-dimensional

Using this dataframe

df = pd.DataFrame({"A": ["foo", "foo", "foo", "foo", "foo",
                         "bar", "bar", "bar", "bar"],
                   "B": ["one", "one", "one", "two", "two",
                         "one", "one", "two", "two"],
                   "C": ["small", "large", "large", "small",
                         "small", "large", "small", "small",
                         "large"],
                   "D": [1, 2, 2, 3, 3, 4, 5, 6, 7],
                   "E": [2, 4, 5, 5, 6, 6, 8, 9, 9]})
'''
     A    B      C  D  E
0  foo  one  small  1  2
1  foo  one  large  2  4
2  foo  one  large  2  5
3  foo  two  small  3  5
4  foo  two  small  3  6
5  bar  one  large  4  6
6  bar  one  small  5  8
7  bar  two  small  6  9
8  bar  two  large  7  9
'''

When I run 

print(pd.pivot_table(df, values='C', index=['A', 'B'],
                    columns=['C'], aggfunc='count'))

To count the number of small/large according to columns A and B (say, for A,B=(foo,one) we have 1 small, and 2 large in column C)

it gives me error

ValueError: Grouper for 'C' not 1-dimensional

What's the problem and how to resolve it?

Upvotes: 5

Views: 13808

Answers (2)

Valdi_Bo
Valdi_Bo

Reputation: 31011

You can not have C column as both values and columns. Probably you should change to:

print(pd.pivot_table(df, index=['A', 'B'], columns=['C'], aggfunc='count'))

Then the result is:

            D           E      
C       large small large small
A   B                          
bar one   1.0   1.0   1.0   1.0
    two   1.0   1.0   1.0   1.0
foo one   2.0   1.0   2.0   1.0
    two   NaN   2.0   NaN   2.0

Upvotes: 6

BallpointBen
BallpointBen

Reputation: 13989

It sounds like what you're after is actually a groupby:

df.groupby(['A', 'B', 'C']).size()

A    B    C    
bar  one  large    1
          small    1
     two  large    1
          small    1
foo  one  large    2
          small    1
     two  small    2
dtype: int64

If you want to then place 'C' back in the columns, you can unstack:

df.groupby(['A', 'B', 'C']).size().unstack().fillna(0)

C        large  small
A   B                
bar one    1.0    1.0
    two    1.0    1.0
foo one    2.0    1.0
    two    0.0    2.0

Upvotes: 2

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