熊猫唯一值多列-Python 实用宝典

熊猫唯一值多列

df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'], 'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'], 'Col3': np.random.random(5)}) 返回“ Col1”和“ Col2”的唯一值的最佳方法是什么? 所需的输出是 'Bob', 'Joe', 'Bill', 'Mary', 'Steve'

问题:熊猫唯一值多列

df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
                   'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
                   'Col3': np.random.random(5)})

返回“ Col1”和“ Col2”的唯一值的最佳方法是什么?

所需的输出是

'Bob', 'Joe', 'Bill', 'Mary', 'Steve'
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'

回答 0

pd.unique 从输入数组或DataFrame列或索引返回唯一值。

此函数的输入必须是一维的,因此将需要合并多列。最简单的方法是选择所需的列,然后在展平的NumPy数组中查看值。整个操作如下所示:

>>> pd.unique(df[['Col1', 'Col2']].values.ravel('K'))
array(['Bob', 'Joe', 'Bill', 'Mary', 'Steve'], dtype=object)

请注意,这ravel()是一个数组方法,它返回多维数组的视图(如果可能)。该参数'K'告诉方法按元素在内存中存储的顺序展平数组(熊猫通常以Fortran连续的顺序存储基础数组;列在行之前)。这可能比使用该方法的默认“ C”顺序快得多。


另一种方法是选择列并将其传递给np.unique

>>> np.unique(df[['Col1', 'Col2']].values)
array(['Bill', 'Bob', 'Joe', 'Mary', 'Steve'], dtype=object)

ravel()此处不需要使用该方法,因为该方法可以处理多维数组。即使这样,它也可能比pd.unique使用基于排序的算法而不是哈希表来标识唯一值的方法要慢。

对于较大的DataFrame,速度上的差异非常大(尤其是在只有少数唯一值的情况下):

>>> 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

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 than 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 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

回答 1

DataFrame在其列中设置了一些简单的字符串:

>>> df
   a  b
0  a  g
1  b  h
2  d  a
3  e  e

您可以连接感兴趣的列并调用unique函数:

>>> pandas.concat([df['a'], df['b']]).unique()
array(['a', 'b', 'd', 'e', 'g', 'h'], dtype=object)

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)

回答 2

In [5]: set(df.Col1).union(set(df.Col2))
Out[5]: {'Bill', 'Bob', 'Joe', 'Mary', 'Steve'}

要么:

set(df.Col1) | set(df.Col2)
In [5]: set(df.Col1).union(set(df.Col2))
Out[5]: {'Bill', 'Bob', 'Joe', 'Mary', 'Steve'}

Or:

set(df.Col1) | set(df.Col2)

回答 3

如果使用多个列,则使用numpy v1.13 +更新的解决方案需要在np.unique中指定轴,否则该数组将隐式展平。

import numpy as np

np.unique(df[['col1', 'col2']], axis=0)

此更改于2016年11月引入:https : //github.com/numpy/numpy/commit/1f764dbff7c496d6636dc0430f083ada9ff4e4be

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: https://github.com/numpy/numpy/commit/1f764dbff7c496d6636dc0430f083ada9ff4e4be


回答 4

pandas解决方案:使用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)

输出:

   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'])

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'])

回答 5

对于那些喜欢大熊猫的人来说,适用于它们,当然还有lambda函数:

df['Col3'] = df[['Col1', 'Col2']].apply(lambda x: ''.join(x), axis=1)

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)

回答 6

这是另一种方式


import numpy as np
set(np.concatenate(df.values))

here's another way


import numpy as np
set(np.concatenate(df.values))

回答 7

list(set(df[['Col1', 'Col2']].as_matrix().reshape((1,-1)).tolist()[0]))

输出将是['Mary','Joe','Steve','Bob','Bill']

list(set(df[['Col1', 'Col2']].as_matrix().reshape((1,-1)).tolist()[0]))

The output will be ['Mary', 'Joe', 'Steve', 'Bob', 'Bill']


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