SQL-like window functions in PANDAS: Row Numbering in Python Pandas Dataframe

I come from a sql background and I use the following data processing step frequently:

1. Partition the table of data by one or more fields
2. For each partition, add a rownumber to each of its rows that ranks the row by one or more other fields, where the analyst specifies ascending or descending

EX:

df = pd.DataFrame({'key1' : ['a','a','a','b','a'],
           'data1' : [1,2,2,3,3],
           'data2' : [1,10,2,3,30]})
df
     data1        data2     key1    
0    1            1         a           
1    2            10        a        
2    2            2         a       
3    3            3         b       
4    3            30        a        

I'm looking for how to do the PANDAS equivalent to this sql window function:

RN = ROW_NUMBER() OVER (PARTITION BY Key1 ORDER BY Data1 ASC, Data2 DESC)


    data1        data2     key1    RN
0    1            1         a       1    
1    2            10        a       2 
2    2            2         a       3
3    3            3         b       1
4    3            30        a       4

I've tried the following which I've gotten to work where there are no 'partitions':

def row_number(frame,orderby_columns, orderby_direction,name):
    frame.sort_index(by = orderby_columns, ascending = orderby_direction, inplace = True)
    frame[name] = list(xrange(len(frame.index)))

I tried to extend this idea to work with partitions (groups in pandas) but the following didn't work:

df1 = df.groupby('key1').apply(lambda t: t.sort_index(by=['data1', 'data2'], ascending=[True, False], inplace = True)).reset_index()

def nf(x):
    x['rn'] = list(xrange(len(x.index)))

df1['rn1'] = df1.groupby('key1').apply(nf)

But I just got a lot of NaNs when I do this.

Ideally, there'd be a succinct way to replicate the window function capability of sql (i've figured out the window based aggregates...that's a one liner in pandas)...can someone share with me the most idiomatic way to number rows like this in PANDAS?

Solution 1:

you can also use sort_values(), groupby() and finally cumcount() + 1:

df['RN'] = df.sort_values(['data1','data2'], ascending=[True,False]) \
             .groupby(['key1']) \
             .cumcount() + 1
print(df)

yields:

   data1  data2 key1  RN
0      1      1    a   1
1      2     10    a   2
2      2      2    a   3
3      3      3    b   1
4      3     30    a   4

PS tested with pandas 0.18

Solution 2:

You can do this by using groupby twice along with the rank method:

In [11]: g = df.groupby('key1')

Use the min method argument to give values which share the same data1 the same RN:

In [12]: g['data1'].rank(method='min')
Out[12]:
0    1
1    2
2    2
3    1
4    4
dtype: float64

In [13]: df['RN'] = g['data1'].rank(method='min')

And then groupby these results and add the rank with respect to data2:

In [14]: g1 = df.groupby(['key1', 'RN'])

In [15]: g1['data2'].rank(ascending=False) - 1
Out[15]:
0    0
1    0
2    1
3    0
4    0
dtype: float64

In [16]: df['RN'] += g1['data2'].rank(ascending=False) - 1

In [17]: df
Out[17]:
   data1  data2 key1  RN
0      1      1    a   1
1      2     10    a   2
2      2      2    a   3
3      3      3    b   1
4      3     30    a   4

It feels like there ought to be a native way to do this (there may well be!...).

Solution 3:

Use groupby.rank function. Here the working example.

df = pd.DataFrame({'C1':['a', 'a', 'a', 'b', 'b'], 'C2': [1, 2, 3, 4, 5]})
df

C1 C2
a  1
a  2
a  3
b  4
b  5

df["RANK"] = df.groupby("C1")["C2"].rank(method="first", ascending=True)
df

C1 C2 RANK
a  1  1
a  2  2
a  3  3
b  4  1
b  5  2

Solution 4:

You can use transform and Rank together Here is an example

df = pd.DataFrame({'C1' : ['a','a','a','b','b'],
           'C2' : [1,2,3,4,5]})
df['Rank'] = df.groupby(by=['C1'])['C2'].transform(lambda x: x.rank())
df

Have a look at Pandas Rank method for more information