在 pandas 中连续获取唯一值 [英] Get count unique values in a row in pandas

问题描述

假设我具有以下数据框:

Suppose I have the following data frame:

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:

How would I get the number of unique (non-NaN) values in a row, such as:

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())) ??

推荐答案

使用列表理解....和set:

Use a list comprehension.... with set:

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进行此操作.

---

You could do this with stack, groupby and nunique.

# 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)

这篇关于在 pandas 中连续获取唯一值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！