从列表中查找数据框中单词的频率

Question

import pandas as pd
list = ['apple','banana','cherries','dragonfruit','elderberry']
data = {'name': ['Alpha', 'Bravo','Charlie','Delta','Echo'],
             'favorite_fruit':  ['apple banana cherries', 'banana cherries dragonfruit',
                           'cherries dragonfruit','dragonfruit','apple elderberry']}

df = pd.DataFrame (data, columns = ['name','favorite_fruit'])

我想计算df中list中每个水果的频率。 预期输出：

df2
Fruit       | Frequency
Apple       |    2     
Banana      |    2
Cherries    |    3
Dragonfruit |    3
Elderberry  |    1

代码 df.favorite_fruit.str.split(expand=True).stack().value_counts() 适用于小型 DataFrame。 如果df.favorite_fruit包含数千行不同的水果组合， 如何仅找到list 中单词的频率？

Answer 1

也许这是一个漏洞答案，但您可以从您已经描述的答案中过滤掉值。所以如果你从这个开始：

>>> df2 = df.favorite_fruit.str.split(expand=True).stack()
>>> df2

0  0          apple
   1         banana
   2       cherries
1  0         banana
   1       cherries
   2    dragonfruit
2  0       cherries
   1    dragonfruit
3  0    dragonfruit
4  0          apple
   1     elderberry
dtype: object

您可以使用isin 将数据限制为目标列表中的数据：

>>> target = ['apple', 'banana']
>>> df2[df2.isin(target)].value_counts()

banana    2
apple     2
dtype: int64

甚至在你原来的答案之后：

>>> df.favorite_fruit.str.split(expand=True).stack().value_counts().loc[target]

apple     2
banana    2
dtype: int64

如果问题是 expand 和 stack 操作对于这么多数据的成本很高，那么这可能不会令人满意。但我认为这可能比基于循环的答案更好？

Answer 2

也许这样做有点迂回，但如果你的favorite_fruit 列总是用空格分隔，这样的事情应该可以工作：

import pandas as pd

list = ['apple','banana','cherries','dragonfruit','elderberry']
data = {'name': ['Alpha', 'Bravo','Charlie','Delta','Echo'],
             'favorite_fruit':  ['apple banana cherries', 'banana cherries dragonfruit',
                           'cherries dragonfruit','dragonfruit','apple elderberry']}

df = pd.DataFrame (data, columns = ['name','favorite_fruit'])

new_df = pd.DataFrame()
data = {}

for i, row in df.iterrows():
    s = row['favorite_fruit']
    items = s.split(' ')
    for item in items:
        if item in data.keys():
            data[item].append(1)
        else:
            data[item] = [1]

for key, value in data.items():
    data[key] = sum(value)

fruit = []
frequency = []

for key, value in data.items():
    fruit.append(key)
    frequency.append(value)

new_df = pd.DataFrame({'fruit': fruit, 'frequency':frequency})
print(new_df)

这会打印出以下内容：

         fruit  frequency
0        apple          2
1       banana          2
2     cherries          3
3  dragonfruit          3
4   elderberry          1

Answer 3

拆分后尝试使用explode功能。

df.favorite_fruit.str.split().explode().value_counts()

cherries       3
dragonfruit    3
banana         2
apple          2
elderberry     1
Name: favorite_fruit, dtype: int64