从 Pandas 数据框中计算不同的单词 [英] Count distinct words from a Pandas Data Frame

问题描述

我有一个 Pandas 数据框，其中一列包含文本.我想获得整个列中出现的唯一单词列表(空格是唯一的分隔符).

I've a Pandas data frame, where one column contains text. I'd like to get a list of unique words appearing across the entire column (space being the only split).

import pandas as pd

r1=['My nickname is ft.jgt','Someone is going to my place']

df=pd.DataFrame(r1,columns=['text'])

输出应如下所示:

['my','nickname','is','ft.jgt','someone','going','to','place']

计数也无妨，但这不是必需的.

It wouldn't hurt to get a count as well, but it is not required.

推荐答案

使用 set 来创建唯一元素的序列.

Use a set to create the sequence of unique elements.

对 df 进行一些清理以获取小写字符串并拆分:

Do some clean-up on df to get the strings in lower case and split:

df['text'].str.lower().str.split()
Out[43]: 
0             [my, nickname, is, ft.jgt]
1    [someone, is, going, to, my, place]

此列中的每个列表都可以传递给 set.update 函数以获得唯一值.使用 apply这样做:

Each list in this column can be passed to set.update function to get unique values. Use apply to do so:

results = set()
df['text'].str.lower().str.split().apply(results.update)
print(results)

set(['someone', 'ft.jgt', 'my', 'is', 'to', 'going', 'place', 'nickname'])

或与评论中的 Counter() 一起使用:

Or use with Counter() from comments:

from collections import Counter
results = Counter()
df['text'].str.lower().str.split().apply(results.update)
print(results)

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