使用 pandas groupby 计算唯一值

本文介绍了使用 pandas groupby 计算唯一值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有以下形式的数据:

df = pd.DataFrame({
    'group': [1, 1, 2, 3, 3, 3, 4],
    'param': ['a', 'a', 'b', np.nan, 'a', 'a', np.nan]
})
print(df)

#    group param
# 0      1     a
# 1      1     a
# 2      2     b
# 3      3   NaN
# 4      3     a
# 5      3     a
# 6      4   NaN

组内的非空值始终相同.我想为每个组(它存在的地方)计算一次非空值，然后找到每个值的总计数.

Non-null values within groups are always the same. I want to count the non-null value for each group (where it exists) once, and then find the total counts for each value.

我目前正在以下列(笨拙且低效的)方式执行此操作:

I'm currently doing this in the following (clunky and inefficient) way:

param = []
for _, group in df[df.param.notnull()].groupby('group'):
    param.append(group.param.unique()[0])
print(pd.DataFrame({'param': param}).param.value_counts())

# a    2
# b    1

我确信有一种方法可以更干净地完成此操作并且不使用循环，但我似乎无法解决.任何帮助将不胜感激.

I'm sure there's a way to do this more cleanly and without using a loop, but I just can't seem to work it out. Any help would be much appreciated.

推荐答案

我认为你可以使用 SeriesGroupBy.nunique:

I think you can use SeriesGroupBy.nunique:

print (df.groupby('param')['group'].nunique())
param
a    2
b    1
Name: group, dtype: int64

另一种使用 的解决方案unique，然后通过，通过和最后一个 value_counts:

Another solution with unique, then create new df by DataFrame.from_records, reshape to Series by stack and last value_counts:

a = df[df.param.notnull()].groupby('group')['param'].unique()
print (pd.DataFrame.from_records(a.values.tolist()).stack().value_counts())
a    2
b    1
dtype: int64

这篇关于使用 pandas groupby 计算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！

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