计算 pandas 数据框中单词的频率

Question

我有一张如下表：

      URN                   Firm_Name
0  104472               R.X. Yah & Co
1  104873        Big Building Society
2  109986          St James's Society
3  114058  The Kensington Society Ltd
4  113438      MMV Oil Associates Ltd

我想计算 Firm_Name 列中所有单词的频率，以获得如下输出：

我尝试过以下代码：

import pandas as pd
import nltk
data = pd.read_csv("X:\Firm_Data.csv")
top_N = 20
word_dist = nltk.FreqDist(data['Firm_Name'])
print('All frequencies')
print('='*60)
rslt=pd.DataFrame(word_dist.most_common(top_N),columns=['Word','Frequency'])

print(rslt)
print ('='*60)

但是，以下代码不会产生唯一的字数统计。

Answer 1

IIUIC，使用

value_counts()

In [3361]: df.Firm_Name.str.split(expand=True).stack().value_counts()
Out[3361]:
Society       3
Ltd           2
James's       1
R.X.          1
Yah           1
Associates    1
St            1
Kensington    1
MMV           1
Big           1
&             1
The           1
Co            1
Oil           1
Building      1
dtype: int64

---

或者，

pd.Series(np.concatenate([x.split() for x in df.Firm_Name])).value_counts()

---

或者，

pd.Series(' '.join(df.Firm_Name).split()).value_counts()

---

对于前 N 个，例如 3

In [3379]: pd.Series(' '.join(df.Firm_Name).split()).value_counts()[:3]
Out[3379]:
Society    3
Ltd        2
James's    1
dtype: int64

---

详情

In [3380]: df
Out[3380]:
      URN                   Firm_Name
0  104472               R.X. Yah & Co
1  104873        Big Building Society
2  109986          St James's Society
3  114058  The Kensington Society Ltd
4  113438      MMV Oil Associates Ltd

Answer 2

您首先需要

str.cat

和

lower

将所有值连接为 1

string

，然后需要

word_tokenize

，最后使用您的解决方案：

top_N = 4
#if not necessary all lower
a = data['Firm_Name'].str.lower().str.cat(sep=' ')
words = nltk.tokenize.word_tokenize(a)
word_dist = nltk.FreqDist(words)
print (word_dist)
<FreqDist with 17 samples and 20 outcomes>

rslt = pd.DataFrame(word_dist.most_common(top_N),
                    columns=['Word', 'Frequency'])
print(rslt)
      Word  Frequency
0  society          3
1      ltd          2
2      the          1
3       co          1

如有必要，也可以删除

lower

：

top_N = 4
a = data['Firm_Name'].str.cat(sep=' ')
words = nltk.tokenize.word_tokenize(a)
word_dist = nltk.FreqDist(words)
rslt = pd.DataFrame(word_dist.most_common(top_N),
                    columns=['Word', 'Frequency'])
print(rslt)
         Word  Frequency
0     Society          3
1         Ltd          2
2         MMV          1
3  Kensington          1

Answer 3

这样更快：

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

Answer 4

这个答案也可以使用 - 从 Pandas 数据框中计算不同的单词。它利用 Counter 方法并将其应用于每一行。

from collections import Counter
c = Counter()
df = pd.DataFrame(
    [[104472,"R.X. Yah & Co"],
    [104873,"Big Building Society"],
    [109986,"St James's Society"],
    [114058,"The Kensington Society Ltd"],
    [113438,"MMV Oil Associates Ltd"]
], columns=["URN","Firm_Name"])
df.Firm_Name.str.split().apply(c.update)

Counter({'R.X.': 1,
         'Yah': 1,
         '&': 1,
         'Co': 1,
         'Big': 1,
         'Building': 1,
         'Society': 3,
         'St': 1,
         "James's": 1,
         'The': 1,
         'Kensington': 1,
         'Ltd': 2,
         'MMV': 1,
         'Oil': 1,
         'Associates': 1})

Answer 5

将整个系列连接成一个字符串，将其拆分为单词，然后使用

Counter

中的

collections

:

from collections import Counter

Counter(df['Firm_Name'].str.cat(sep='\n').split())

输出：

Counter({'Society': 3,
         'Ltd': 2,
         'R.X.': 1,
         'Yah': 1,
         '&': 1,
         'Co': 1,
         'Big': 1,
         'Building': 1,
         'St': 1,
         "James's": 1,
         'The': 1,
         'Kensington': 1,
         'MMV': 1,
         'Oil': 1,
         'Associates': 1})

可以选择调用

.most_common()

来获取元组的排序列表

[(w1, c1), (w2, c2), ...]