Set Multi-Index DataFrame column by Series with Index - python

I'm struggling with a MultiIndex dataframe (a) which requires the column x to be set by b which isn't a MultiIndex and has only 1 index level (first level of a). I have an index to change those values (ix), which is why I am using .loc[] for indexing. The problem is that the way missing index levels are populated in a is not what I require (see example).
>>> a = pd.DataFrame({'a': [1, 2, 3], 'b': ['b', 'b', 'b'], 'x': [4, 5, 6]}).set_index(['a', 'b'])
>>> a
x
a b
1 b 4
2 b 5
3 b 6
>>> b = pd.DataFrame({'a': [1, 4], 'x': [9, 10]}).set_index('a')
>>> b
x
a
1 9
4 10
>>> ix = a.index[[0, 1]]
>>> ix
MultiIndex(levels=[[1, 2, 3], [u'b']],
codes=[[0, 1], [0, 0]],
names=[u'a', u'b'])
>>> a.loc[ix]
x
a b
1 b 4
2 b 5
>>> a.loc[ix, 'x'] = b['x']
>>> # wrong result (at least not what I want)
>>> a
x
a b
1 b NaN
2 b NaN
3 b 6.0
>>> # expected result
>>> a
x
a b
1 b 9 # index: a=1 is part of DataFrame b
2 b 5 # other indices don't exist in b and...
3 b 6 # ... x-values remain unchanged
# if there were more [1, ...] indices...
# ...x would also bet set to 9

I think you want to merge a and B. you should consider using concat,merge or join funcs.

I can't think of any one-liner, so here's a multi-step approach:
tmp_df = a.loc[ix, ['x']].reset_index(level=1, drop=True)
tmp_df['x'] = b['x']
tmp_df.index = ix
a.loc[ix, 'x'] = tmp_df['x']
Output:
x
a b
1 b 9.0
2 b 5.0
3 b 6.0
Edit: I assume that the b's in index are symbolic. Otherwise, the code will fail from a.loc[ix, 'x']: for
a = pd.DataFrame({'a': [1, 1, 2, 3],
'b': ['b', 'b', 'b', 'b'],
'x': [4, 5, 3, 6]}).set_index(['a', 'b'])
a.loc[ix,'x'] gives:
a b
1 b 4
b 5
b 4
b 5
Name: x, dtype: int64

You try use 1- index frame with 2- index frame, just use values:
EDIT:
import pandas as pd
a = pd.DataFrame({'a': [1, 2, 3], 'b': ['b', 'b', 'b'], 'x': [4, 5, 6]}).set_index(['a', 'b'])
b = pd.DataFrame({'a': [1, 4], 'x': [9, 10]}).set_index('a')
a_ix = a.index.get_level_values('a')[[0, 1]]
b_ix = b.index
mask = (b_ix == a_ix)
a.loc[mask, 'x'] = b.loc[mask,'x'].values
a:
x
a b
1 b 9
2 b 5
3 b 6

I first reset the multi-index of a and then I set it to the (single column) a
a = a.reset_index()
a = a.set_index('a')
print(a)
b x
a
1 b 4
2 b 5
3 b 6
print(b)
x
a
1 9
4 10
Then, make the assignment you require using loc and also re-set the multi-index
now, since we are using loc, your ix = a.index[[0, 1]] becomes similar to [1,0] (1 refers to index of a and 0 refers to index of b)
a.loc[1, 'x'] = b.iloc[0,0]
a.reset_index(inplace=True)
a = a.set_index(['a','b'])
print(a)
x
a b
1 b 9
2 b 5
3 b 6
EDIT:
Alternatively, reset the multi-index of a and don't set it to a single column index. Then your [0,1] (referring to index values with loc, not positions iloc) can be used (0 refers to index of a and 1 refers to index of b)
a = a.reset_index()
print(a)
a b x
0 1 b 4
1 2 b 5
2 3 b 6
a.loc[0, 'x'] = b.loc[1,'x']
a = a.set_index(['a','b'])
print(a)
x
a b
1 b 9
2 b 5
3 b 6

Related

Pandas, how to pick value from different columns based on value from diffrent column? [duplicate]

The operation pandas.DataFrame.lookup is "Deprecated since version 1.2.0", and has since invalidated a lot of previous answers.
This post attempts to function as a canonical resource for looking up corresponding row col pairs in pandas versions 1.2.0 and newer.
Standard LookUp Values With Default Range Index
Given the following DataFrame:
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
Col A B
0 B 1 5
1 A 2 6
2 A 3 7
3 B 4 8
I would like to be able to lookup the corresponding value in the column specified in Col:
I would like my result to look like:
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
Standard LookUp Values With a Non-Default Index
Non-Contiguous Range Index
Given the following DataFrame:
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
Col A B
0 B 1 5
2 A 2 6
8 A 3 7
9 B 4 8
I would like to preserve the index but still find the correct corresponding Value:
Col A B Val
0 B 1 5 5
2 A 2 6 2
8 A 3 7 3
9 B 4 8 8
MultiIndex
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=pd.MultiIndex.from_product([['C', 'D'], ['E', 'F']]))
Col A B
C E B 1 5
F A 2 6
D E A 3 7
F B 4 8
I would like to preserve the index but still find the correct corresponding Value:
Col A B Val
C E B 1 5 5
F A 2 6 2
D E A 3 7 3
F B 4 8 8
LookUp with Default For Unmatched/Not-Found Values
Given the following DataFrame
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'C'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
Col A B
0 B 1 5
1 A 2 6
2 A 3 7
3 C 4 8 # Column C does not correspond with any column
I would like to look up the corresponding values if one exists otherwise I'd like to have it default to 0
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 C 4 8 0 # Default value 0 since C does not correspond
LookUp with Missing Values in the lookup Col
Given the following DataFrame:
Col A B
0 B 1 5
1 A 2 6
2 A 3 7
3 NaN 4 8 # <- Missing Lookup Key
I would like any NaN values in Col to result in a NaN value in Val
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 A 3 7 3.0
3 NaN 4 8 NaN # NaN to indicate missing
Standard LookUp Values With Any Index
The documentation on Looking up values by index/column labels recommends using NumPy indexing via factorize and reindex as the replacement for the deprecated DataFrame.lookup.
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
factorize is used to convert the column encode the values as an "enumerated type".
idx, col = pd.factorize(df['Col'])
# idx = array([0, 1, 1, 0], dtype=int64)
# col = Index(['B', 'A'], dtype='object')
Notice that B corresponds to 0 and A corresponds to 1. reindex is used to ensure that columns appear in the same order as the enumeration:
df.reindex(columns=col)
B A # B appears First (location 0) A appers second (location 1)
0 5 1
1 6 2
2 7 3
3 8 4
We need to create an appropriate range indexer compatible with NumPy indexing.
The standard approach is to use np.arange based on the length of the DataFrame:
np.arange(len(df))
[0 1 2 3]
Now NumPy indexing will work to select values from the DataFrame:
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
[5 2 3 8]
*Note: This approach will always work regardless of type of index.
MultiIndex
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=pd.MultiIndex.from_product([['C', 'D'], ['E', 'F']]))
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
Col A B Val
C E B 1 5 5
F A 2 6 2
D E A 3 7 3
F B 4 8 8
Why use np.arange and not df.index directly?
Standard Contiguous Range Index
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
In this case only, there is no error as the result from np.arange is the same as the df.index.
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
Non-Contiguous Range Index Error
Raises IndexError:
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
IndexError: index 8 is out of bounds for axis 0 with size 4
MultiIndex Error
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=pd.MultiIndex.from_product([['C', 'D'], ['E', 'F']]))
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
Raises IndexError:
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices
LookUp with Default For Unmatched/Not-Found Values
There are a few approaches.
First let's look at what happens by default if there is a non-corresponding value:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'C'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
# Col A B
# 0 B 1 5
# 1 A 2 6
# 2 A 3 7
# 3 C 4 8
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 A 3 7 3.0
3 C 4 8 NaN # NaN Represents the Missing Value in C
If we look at why the NaN values are introduced, we will find that when factorize goes through the column it will enumerate all groups present regardless of whether they correspond to a column or not.
For this reason, when we reindex the DataFrame we will end up with the following result:
idx, col = pd.factorize(df['Col'])
df.reindex(columns=col)
idx = array([0, 1, 1, 2], dtype=int64)
col = Index(['B', 'A', 'C'], dtype='object')
df.reindex(columns=col)
B A C
0 5 1 NaN
1 6 2 NaN
2 7 3 NaN
3 8 4 NaN # Reindex adds the missing column with the Default `NaN`
If we want to specify a default value, we can specify the fill_value argument of reindex which allows us to modify the behaviour as it relates to missing column values:
idx, col = pd.factorize(df['Col'])
df.reindex(columns=col, fill_value=0)
idx = array([0, 1, 1, 2], dtype=int64)
col = Index(['B', 'A', 'C'], dtype='object')
df.reindex(columns=col, fill_value=0)
B A C
0 5 1 0
1 6 2 0
2 7 3 0
3 8 4 0 # Notice reindex adds missing column with specified value `0`
This means that we can do:
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(
columns=col,
fill_value=0 # Default value for Missing column values
).to_numpy()[np.arange(len(df)), idx]
df:
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 C 4 8 0
*Notice the dtype of the column is int, since NaN was never introduced, and, therefore, the column type was not changed.
LookUp with Missing Values in the lookup Col
factorize has a default na_sentinel=-1, meaning that when NaN values appear in the column being factorized the resulting idx value is -1
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', np.nan],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
# Col A B
# 0 B 1 5
# 1 A 2 6
# 2 A 3 7
# 3 NaN 4 8 # <- Missing Lookup Key
idx, col = pd.factorize(df['Col'])
# idx = array([ 0, 1, 1, -1], dtype=int64)
# col = Index(['B', 'A'], dtype='object')
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
# Col A B Val
# 0 B 1 5 5
# 1 A 2 6 2
# 2 A 3 7 3
# 3 NaN 4 8 4 <- Value From A
This -1 means that, by default, we'll be pulling from the last column when we reindex. Notice the col still only contains the values B and A. Meaning, that we will end up with the value from A in Val for the last row.
The easiest way to handle this is to fillna Col with some value that cannot be found in the column headers.
Here I use the empty string '':
idx, col = pd.factorize(df['Col'].fillna(''))
# idx = array([0, 1, 1, 2], dtype=int64)
# col = Index(['B', 'A', ''], dtype='object')
Now when I reindex, the '' column will contain NaN values meaning that the lookup produces the desired result:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', np.nan],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
idx, col = pd.factorize(df['Col'].fillna(''))
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
df:
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 A 3 7 3.0
3 NaN 4 8 NaN # Missing as expected
Other Approaches to LookUp
There are 2 other approaches to performing this operation:
apply (Intuitive, but quite slow)
apply can be used on axis=1 in order to use the Column values as the key:
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
df['Val'] = df.apply(lambda row: row[row['Col']], axis=1)
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
This operation will work regardless of index type:
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
# Col A B
# 0 B 1 5
# 2 A 2 6
# 8 A 3 7
# 9 B 4 8
df['Val'] = df.apply(lambda row: row[row['Col']], axis=1)
df:
Col A B Val
0 B 1 5 5
2 A 2 6 2
8 A 3 7 3
9 B 4 8 8
When dealing with Missing/Non-Corresponding Values we can use Series.get can be used to remedy this issue:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'C', np.nan],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
# Col A B
# 0 B 1 5
# 1 A 2 6
# 2 C 3 7 <- Non Corresponding
# 3 NaN 4 8 <- Missing
df['Val'] = df.apply(lambda row: row.get(row['Col']), axis=1)
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 C 3 7 NaN # Missing value
3 NaN 4 8 NaN # Missing value
With Default Value
df['Val'] = df.apply(lambda row: row.get(row['Col'], default=-1), axis=1)
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 C 3 7 -1 # Default -1
3 NaN 4 8 -1 # Default -1
apply is extremely flexible and modifications are straightforward, however, the general iterative approach, as well as all the individual Series lookups can become extremely costly in large DataFrames.
get_indexer (limited)
Index.get_indexer can be used to convert the column to index values into an indexer for the DataFrame. This means there is no reason to reindex the DataFrame as the indexer corresponds to the DataFrame as a whole.
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
df['Val'] = df.to_numpy()[df.index, df.columns.get_indexer(df['Col'])]
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
This approach is reasonably fast, however, missing values are represented by -1 meaning that if a value is missing it will grab the value from the -1 column (The last column in the DataFrame).
import pandas as pd
df = pd.DataFrame({'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8],
'Col': ['B', 'A', 'A', 'C']})
# A B Col <- Col is now the Last Col
# 0 1 5 B
# 1 2 6 A
# 2 3 7 A
# 3 4 8 C <- Notice Col `C` does not correspond to a Valid Column Header
df['Val'] = df.to_numpy()[df.index, df.columns.get_indexer(df['Col'])]
df:
A B Col Val
0 1 5 B 5
1 2 6 A 2
2 3 7 A 3
3 4 8 C C # <- Value from the last column in the DataFrame (index -1)
It is also notable that not reindexing the DataFrame means converting the entire DataFrame to numpy. This can be very costly if there are many unrelated columns that all need converted:
import numpy as np
import pandas as pd
df = pd.DataFrame({1: 10,
2: 20,
3: 't',
4: 40,
5: np.nan,
'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
df['Val'] = df.to_numpy()[df.index, df.columns.get_indexer(df['Col'])]
df.to_numpy()
[[10 20 't' 40 nan 'B' 1 5 5]
[10 20 't' 40 nan 'A' 2 6 2]
[10 20 't' 40 nan 'A' 3 7 3]
[10 20 't' 40 nan 'B' 4 8 8]]
Compared to the reindexing approach which only contains columns relevant to the column values:
df.reindex(columns=['B', 'A']).to_numpy()
[[5 1]
[6 2]
[7 3]
[8 4]]
Another option is to build a tuple of the lookup columns, pivot the dataframe, and select the relevant columns with the tuples:
cols = [(ent, ent) for ent in df.Col.unique()]
df.assign(Val = df.pivot(index = None, columns = 'Col')
.reindex(columns = cols)
.ffill(axis=1)
.iloc[:, -1])
Col A B Val
0 B 1 5 5.0
2 A 2 6 2.0
8 A 3 7 3.0
9 B 4 8 8.0
Another possible method is to use melt:
df['value'] = (df.melt('Col', ignore_index=False)
.loc[lambda x: x['Col'] == x['variable'], 'value'])
print(df)
# Output:
Col A B value
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
This method also works with Missing/Non-Corresponding Values:
df['value'] = (df.melt('Col', ignore_index=False)
.loc[lambda x: x['Col'] == x['variable'], 'value'])
print(df)
# Output
Col A B value
0 B 1 5 5.0
1 A 2 6 2.0
2 C 3 7 NaN
3 NaN 4 8 NaN
You can replace .loc[...] by query(...) but it's little slower although more expressive:
df['value'] = df.melt('Col', ignore_index=False).query('Col == variable')['value']

pandas reorder only a specific row

I have a Dataframe file in which I want to switch the order of columns in only the third row while keeping other rows the same.
Under some condition, I have to switch orders for my project, but here is an example that probably has no real meaning.
Suppose the dataset is
df = pd.DataFrame({'A': [0, 1, 2, 3, 4],
'B': [5, 6, 7, 8, 9],
'C': ['a', 'b', 'c', 'd', 'e']})
df
out[1]:
A B C
0 0 5 a
1 1 6 b
2 2 7 c
3 3 8 d
4 4 9 e
I want to have the output:
A B C
0 0 5 a
1 1 6 b
2 **7 2** c
3 3 8 d
4 4 9 e
How do I do it?
I have tried:
new_order = [1, 0, 2] # specify new order of the third row
i = 2 # specify row number
df.iloc[i] = df[df.columns[new_order]].loc[i] # reorder the third row only and assign new values to df
I observed from the output of the right-hand side that the columns are reordering as I wanted:
df[df.columns[new_order]].loc[i]
Out[2]:
B 7
A 2
C c
Name: 2, dtype: object
But when assigned to df again, it did nothing. I guess it's because of the name matching.
Can someone help me? Thanks in advance!

Create a column in a dataframe that is a string of characters summarizing data in other columns

I have a dataframe like this where the columns are the scores of some metrics:
A B C D
4 3 3 1
2 5 2 2
3 5 2 4
I want to create a new column to summarize which metrics each row scored over a set threshold in, using the column name as a string. So if the threshold was A > 2, B > 3, C > 1, D > 3, I would want the new column to look like this:
A B C D NewCol
4 3 3 1 AC
2 5 2 2 BC
3 5 2 4 ABCD
I tried using a series of np.where:
df[NewCol] = np.where(df['A'] > 2, 'A', '')
df[NewCol] = np.where(df['B'] > 3, 'B', '')
etc.
but realized the result was overwriting with the last metric any time all four metrics didn't meet the conditions, like so:
A B C D NewCol
4 3 3 1 C
2 5 2 2 C
3 5 2 4 ABCD
I am pretty sure there is an easier and correct way to do this.
You could do:
import pandas as pd
data = [[4, 3, 3, 1],
[2, 5, 2, 2],
[3, 5, 2, 4]]
df = pd.DataFrame(data=data, columns=['A', 'B', 'C', 'D'])
th = {'A': 2, 'B': 3, 'C': 1, 'D': 3}
df['result'] = [''.join(k for k in df.columns if record[k] > th[k]) for record in df.to_dict('records')]
print(df)
Output
A B C D result
0 4 3 3 1 AC
1 2 5 2 2 BC
2 3 5 2 4 ABCD
Using dot
s=pd.Series([2,3,1,3],index=df.columns)
df.gt(s,1).dot(df.columns)
Out[179]:
0 AC
1 BC
2 ABCD
dtype: object
#df['New']=df.gt(s,1).dot(df.columns)
Another option that operates in an array fashion. It would be interesting to compare performance.
import pandas as pd
import numpy as np
# Data to test.
data = pd.DataFrame(
[
[4, 3, 3, 1],
[2, 5, 2, 2],
[3, 5, 2, 4]
]
, columns = ['A', 'B', 'C', 'D']
)
# Series to hold the thresholds.
thresholds = pd.Series([2, 3, 1, 3], index = ['A', 'B', 'C', 'D'])
# Subtract the series from the data, broadcasting, and then use sum to concatenate the strings.
data['result'] = np.where(data - thresholds > 0, data.columns, '').sum(axis = 1)
print(data)
Gives:
A B C D result
0 4 3 3 1 AC
1 2 5 2 2 BC
2 3 5 2 4 ABCD

Unusual reshaping of Pandas DataFrame

i have a DF like this:
df = pd.DataFrame({'x': ['a', 'a', 'b', 'b', 'b', 'c'],
'y': [1, 2, 3, 4, 5, 6],
})
which looks like:
x y
0 a 1
1 a 2
2 b 3
3 b 4
4 b 5
5 c 6
I need to reshape it in the way to keep 'x' column unique:
x y_1 y_2 y_3
0 a 1 2 NaN
1 b 3 4 5
2 c 6 NaN NaN
So the max N of 'y_N' columns have to be equal to
max(df.groupby('x').count().values)
and the x column has to contain unique values.
For now i dont get how to get y_N columns.
Thanks.
You can use pandas.crosstab with cumcount column as the columns parameter:
(pd.crosstab(df.x, df.groupby('x').cumcount() + 1, df.y,
aggfunc = lambda x: x.iloc[0])
.rename(columns="y_{}".format).reset_index())

Change pivot table from Series to DataFrame

df = pd.DataFrame({'A': [1, 1, 1, 2, 2, 2, 3, 3, 3],
'B': ['X', 'Y', 'Z'] * 3,
'C': [1, 2, 3, 1, 2, 3, 1, 2, 3]})
>>> df
A B C
0 1 X 1
1 1 Y 2
2 1 Z 3
3 2 X 1
4 2 Y 2
5 2 Z 3
6 3 X 1
7 3 Y 2
8 3 Z 3
result = df.pivot_table(index=['B'], values='C', aggfunc=sum)
>>> result
B
X 3
Y 6
Z 9
Name: C, dtype: int64
How can I have the column name for C show up a above the sums, and how can I sort result either ascending or descending. Result is a series not a dataframe and seems non-sortable?
Python: 2.7.11 and Pandas: 0.17.1
You were very close. Note that the brackets around the values coerces the result into a dataframe instead of a series (i.e. values=['C'] instead of values='C').
result = df.pivot_table(index = ['B'], values=['C'], aggfunc=sum)
>>> result
C
B
X 3
Y 6
Z 9
Asresult is now a dataframe, you can use sort_values on it:
>>> result.sort_values('C', ascending=False)
C
B
Z 9
Y 6
X 3

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