I come from a sql background and I use the following data processing step frequently:
- Partition the table of data by one or more fields
- 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?
Answers:
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Method 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
Method 2
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
Method 3
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!…).
Method 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
Method 5
pandas.lib.fast_zip() can create a tuple array from a list of array. You can use this function to create a tuple series, and then rank it:
values = {'key1' : ['a','a','a','b','a','b'],
'data1' : [1,2,2,3,3,3],
'data2' : [1,10,2,3,30,20]}
df = pd.DataFrame(values, index=list("abcdef"))
def rank_multi_columns(df, cols, **kw):
data = []
for col in cols:
if col.startswith("-"):
flag = -1
col = col[1:]
else:
flag = 1
data.append(flag*df[col])
values = pd.lib.fast_zip(data)
s = pd.Series(values, index=df.index)
return s.rank(**kw)
rank = df.groupby("key1").apply(lambda df:rank_multi_columns(df, ["data1", "-data2"]))
print rank
the result:
a 1 b 2 c 3 d 2 e 4 f 1 dtype: float64
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0
