df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
'Col3': np.random.random(5)})
What is the best way to return the unique values of 'Col1' and 'Col2'?
The desired output is
'Bob', 'Joe', 'Bill', 'Mary', 'Steve'
8 Answers
pd.unique returns the unique values from an input array, or DataFrame column or index.
The input to this function needs to be one-dimensional, so multiple columns will need to be combined. The simplest way is to select the columns you want and then view the values in a flattened NumPy array. The whole operation looks like this:
>>> pd.unique(df[['Col1', 'Col2']].values.ravel('K'))
array(['Bob', 'Joe', 'Bill', 'Mary', 'Steve'], dtype=object)
Note that ravel() is an array method that returns a view (if possible) of a multidimensional array. The argument 'K' tells the method to flatten the array in the order the elements are stored in the memory (pandas typically stores underlying arrays in Fortran-contiguous order; columns before rows). This can be significantly faster than using the method's default 'C' order.
An alternative way is to select the columns and pass them to np.unique:
>>> np.unique(df[['Col1', 'Col2']].values)
array(['Bill', 'Bob', 'Joe', 'Mary', 'Steve'], dtype=object)
There is no need to use ravel() here as the method handles multidimensional arrays. Even so, this is likely to be slower than pd.unique as it uses a sort-based algorithm rather than a hashtable to identify unique values.
The difference in speed is significant for larger DataFrames (especially if there are only a handful of unique values):
>>> df1 = pd.concat([df]*100000, ignore_index=True) # DataFrame with 500000 rows
>>> %timeit np.unique(df1[['Col1', 'Col2']].values)
1 loop, best of 3: 1.12 s per loop
>>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel('K'))
10 loops, best of 3: 38.9 ms per loop
>>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel()) # ravel using C order
10 loops, best of 3: 49.9 ms per loop
I have setup a DataFrame with a few simple strings in it's columns:
>>> df
a b
0 a g
1 b h
2 d a
3 e e
You can concatenate the columns you are interested in and call unique function:
>>> pandas.concat([df['a'], df['b']]).unique()
array(['a', 'b', 'd', 'e', 'g', 'h'], dtype=object)
In [5]: set(df.Col1).union(set(df.Col2))
Out[5]: {'Bill', 'Bob', 'Joe', 'Mary', 'Steve'}
Or:
set(df.Col1) | set(df.Col2)
An updated solution using numpy v1.13+ requires specifying the axis in np.unique if using multiple columns, otherwise the array is implicitly flattened.
import numpy as np
np.unique(df[['col1', 'col2']], axis=0)
This change was introduced Nov 2016:
for those of us that love all things pandas, apply, and of course lambda functions:
df['Col3'] = df[['Col1', 'Col2']].apply(lambda x: ''.join(x), axis=1)
here's another way
import numpy as np
set(np.concatenate(df.values))
Non-pandas solution: using set().
import pandas as pd
import numpy as np
df = pd.DataFrame({'Col1' : ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
'Col2' : ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
'Col3' : np.random.random(5)})
print df
print set(df.Col1.append(df.Col2).values)
Output:
Col1 Col2 Col3
0 Bob Joe 0.201079
1 Joe Steve 0.703279
2 Bill Bob 0.722724
3 Mary Bob 0.093912
4 Joe Steve 0.766027
set(['Steve', 'Bob', 'Bill', 'Joe', 'Mary'])
list(set(df[['Col1', 'Col2']].as_matrix().reshape((1,-1)).tolist()[0]))
The output will be ['Mary', 'Joe', 'Steve', 'Bob', 'Bill']
