连续获取熊猫中唯一的计数值

Question

假设我具有以下数据框：

0     1        2
new   NaN      NaN
new   one      one
a     b        c
NaN   NaN      NaN

如何获取一行中唯一（非NaN）值的数量，例如：

0     1        2       _num_unique_values
new   NaN      NaN     1
new   one      one     2
a     b        c       3
NaN   NaN      NaN     0

我想可能是这样的：

df['_num_unique_values'] = len(set(df.loc.tolist())) ??

Answer 1

对set使用列表理解...。

df['num_uniq'] = [len(set(v[pd.notna(v)].tolist())) for v in df.values]
df

     0    1    2  num_uniq
0  new  NaN  NaN         1
1  new  one  one         2
2    a    b    c         3
3  NaN  NaN  NaN         0

---

您可以使用stack，groupby和nunique来做到这一点。

# df.join(df.stack().groupby(level=0).nunique().to_frame('num_uniq'))
df['num_uniq'] = df.stack().groupby(level=0).nunique()
df

     0    1    2  num_uniq
0  new  NaN  NaN       1.0
1  new  one  one       2.0
2    a    b    c       3.0
3  NaN  NaN  NaN       NaN

---

另一个选择是apply和nunique：

df['num_uniq'] = df.apply(pd.Series.nunique, axis=1)
df

     0    1    2  num_uniq
0  new  NaN  NaN         1
1  new  one  one         2
2    a    b    c         3
3  NaN  NaN  NaN         0

---

性能

df_ = df
df = pd.concat([df_] * 1000, ignore_index=True)

%timeit df['num_uniq'] = [len(set(v[pd.notna(v)])) for v in df.values]
%timeit df['num_uniq'] = df.stack().groupby(level=0).nunique()
%timeit df['num_uniq'] = df.apply(pd.Series.nunique, axis=1)
%timeit df['num_uniq'] = df.nunique(1)

196 ms ± 10.1 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
6.34 ms ± 343 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
679 ms ± 24 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
3.21 ms ± 343 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

只需使用nunique（axis = 1）。

import numpy as np
import pandas as pd

data={0:['new','new','a',np.nan],
     1:[np.nan,'one','b', np.nan],
     2:[np.nan,np.nan,'c',np.nan]}
df = pd.DataFrame(data)

print(df.nunique(axis=1))

df['num_unique'] = df.nunique(axis=1)

请参阅：

Answer 3

更抽象的解决方案：

df['num_uniq']=df.nunique(axis=1)

Answer 4

它的速度不及set()的Coldspeed回答，但您也可以这样做

df['_num_unique_values'] = df.T.nunique()

首先，将df数据帧的转置与df.T一起使用，然后使用nunique()获取不包含NaN的唯一值的计数。

这是作为新列添加到原始数据框的。

df现在应该是

    0   1   2   _num_unique_values
0   new nan nan 1
1   new one one 2
2   a   b   c   3
3   nan nan nan 0