Question

我有这样的数据集：

   Type Word

0   N   Work
1   N   Rock
2   N   Rock
3   Adj Rock
4   V   Rock
5   N   Work
6   V   Work
7   V   Rock
8   Adj Like
9   N   Rock
10  V   Love
11  V   Like
12  V   Rock
13  Adj Blue
14  Adv Work

我想计算每个单词的数量，并获得每个单词的前2种类型。我期望的结果如下：

    Word    Top    Count

0   Rock    N, V    7
1   Work    N, Adv  4
2   Like    Adj, V  2
3   Blue    Adj     1
4   Love    V       1

我创建了一些代码行，并按照我的预期得到了结果。这是我的代码：

In [1]: 
import pandas as pd
df = pd.DataFrame([
    ['N','Work'],
    ['N','Rock'],
    ['N','Rock'],
    ['Adj','Rock'], 
    ['V','Rock'],
    ['N','Work'],
    ['V','Work'],
    ['V','Rock'],
    ['Adj','Like'],
    ['N','Rock'],
    ['V','Love'],
    ['V','Like'],
    ['V','Rock'],
    ['Adj','Blue'],
    ['Adv','Work']], columns=['Type', 'Word'])

In [2]: #Group by column "Word","Type" and count number of each pair
df = df.groupby(["Type", "Word"])["Type"].count().reset_index(name="Count")

In [3]:
df
   Type Word    Count
0   Adj Blue    1
1   Adj Like    1
2   Adj Rock    1
3   Adv Work    1
4   N   Rock    3
5   N   Work    2
6   V   Like    1
7   V   Love    1
8   V   Rock    3
9   V   Work    1

In [4]: #Group by "Word" and sort by "Count" in each group, get top 2
df1 = df.sort_values(["Word","Count"], ascending=False).groupby("Word").head(2)
df1
   Type Word    Count
5   N   Work    2
3   Adv Work    1
4   N   Rock    3
8   V   Rock    3
7   V   Love    1
1   Adj Like    1
6   V   Like    1
0   Adj Blue    1

In [5]: #Groupby "Word" and union "Type" in each group
df1 = df1.groupby('Word')['Type'].apply(lambda x: "%s" % ', '.join(x)).reset_index(name='Top')
df1
    Word    Top
0   Blue    Adj
1   Like    Adj, V
2   Love    V
3   Rock    N, V
4   Work    N, Adv

In [6]: #Compute number of each word, save to a new dataframe
df_sum = df.groupby('Word').sum().reset_index()
df_sum
    Word    Count
0   Blue    1
1   Like    2
2   Love    1
3   Rock    7
4   Work    4

In [7]: #Merge to dataframe containing number of each word
df1.merge(df_sum).sort_values("Count", ascending=False)
df1
    Word    Top     Count
3   Rock    N, V    7
4   Work    N, Adv  4
1   Like    Adj, V  2
0   Blue    Adj     1
2   Love    V       1

但是，这段代码似乎不是最优的。我使用了很多groupby，并使用了sort_values 2次。如果数据集实际很大，那将会很麻烦。你能优化它吗？感谢。

Answer 1

df.groupby('Word').agg(dict(
        Type=lambda x: ', '.join(pd.value_counts(x).index[:2]),
        Word='size'
    )).rename(columns=dict(Word='Count')).reset_index().sort_values('Count')

   Word    Type  Count
0  Blue     Adj      1
2  Love       V      1
1  Like  V, Adj      2
4  Work    N, V      4
3  Rock    N, V      7

Answer 2

您可以使用agg后面的Counter来获取最常见的类型，并使用len来计算出现的单词数量。

import pandas as pd
from collections import Counter    

group_df = df.groupby('Word')
df_summary = group_df.agg(
    lambda x: {'Type': [', '.join([e[0] for e in Counter(x.Type).most_common(2)]), len(x)]}
)
df_out = df_summary.Type.apply(pd.Series).reset_index().rename(columns={0: 'Top', 1: 'count'})
df_out.sort_values('count', ascending=False) # output

这将输出数据帧为

    Word    Top count
3   Rock    N, V    7
4   Work    N, V    4
1   Like    Adj, V  2
0   Blue    Adj 1
2   Love    V   1

优化groupby聚合熊猫

2 个答案: