问题
The pandas factorize
function assigns each unique value in a series to a sequential, 0-based index, and calculates which index each series entry belongs to.
I'd like to accomplish the equivalent of pandas.factorize
on multiple columns:
import pandas as pd
df = pd.DataFrame({'x': [1, 1, 2, 2, 1, 1], 'y':[1, 2, 2, 2, 2, 1]})
pd.factorize(df)[0] # would like [0, 1, 2, 2, 1, 0]
That is, I want to determine each unique tuple of values in several columns of a data frame, assign a sequential index to each, and compute which index each row in the data frame belongs to.
Factorize
only works on single columns. Is there a multi-column equivalent function in pandas?
回答1:
You need to create a ndarray of tuple first, pandas.lib.fast_zip
can do this very fast in cython loop.
import pandas as pd
df = pd.DataFrame({'x': [1, 1, 2, 2, 1, 1], 'y':[1, 2, 2, 2, 2, 1]})
print pd.factorize(pd.lib.fast_zip([df.x, df.y]))[0]
the output is:
[0 1 2 2 1 0]
回答2:
I am not sure if this is an efficient solution. There might be better solutions for this.
arr=[] #this will hold the unique items of the dataframe
for i in df.index:
if list(df.iloc[i]) not in arr:
arr.append(list(df.iloc[i]))
so printing the arr would give you
>>>print arr
[[1,1],[1,2],[2,2]]
to hold the indices, i would declare an ind array
ind=[]
for i in df.index:
ind.append(arr.index(list(df.iloc[i])))
printing ind would give
>>>print ind
[0,1,2,2,1,0]
回答3:
You can use drop_duplicates
to drop those duplicated rows
In [23]: df.drop_duplicates()
Out[23]:
x y
0 1 1
1 1 2
2 2 2
EDIT
To achieve your goal, you can join your original df to the drop_duplicated one:
In [46]: df.join(df.drop_duplicates().reset_index().set_index(['x', 'y']), on=['x', 'y'])
Out[46]:
x y index
0 1 1 0
1 1 2 1
2 2 2 2
3 2 2 2
4 1 2 1
5 1 1 0
回答4:
df = pd.DataFrame({'x': [1, 1, 2, 2, 1, 1], 'y':[1, 2, 2, 2, 2, 1]})
tuples = df[['x', 'y']].apply(tuple, axis=1)
df['newID'] = pd.factorize( tuples )[0]
来源:https://stackoverflow.com/questions/16453465/multi-column-factorize-in-pandas