Python 数据帧中特定关键字的计数

我有这样一个数据帧：

    A
0   Please wait outside of the house
1   A glittering gem is not enough.
2   The memory we used to share is no longer coher...
3   She only paints with bold colors; she does not...

我有一组关键词：

keywords = ["of","is","she"]

如何为每个关键字创建一列，其中包含该关键字在dataframe的每个句子中出现的次数？它看起来像：

                                                   A  of  is  she
0                   Please wait outside of the house   1   0    0
1                    A glittering gem is not enough.   0   1    0
2  The memory we used to share is no longer coher...   0   1    0
3  She only paints with bold colors; she does not...   0   0    2

---

注意：我查看了，但它没有回答我的问题。

我假设您正在查找不区分大小写的匹配项

将熊猫作为pd导入
df=pd.DataFrame({
“A”：[
“请在屋外等候”，
“光有闪闪发光的宝石是不够的。”，
“我们过去共享的记忆不再连贯…”，
“她只画大胆的颜色；她不……”
]
})
关键词=[“of”，“is”，“she”]
对于关键字中的关键字：
df[keyword]=df['A'].apply（lambda\u str:\u str.lower（）.count（关键字））
打印（df）

输出

                                                   A  of  is  she
0                   Please wait outside of the house   1   0    0
1                    A glittering gem is not enough.   0   1    0
2  The memory we used to share is no longer coher...   0   1    0
3  She only paints with bold colors; she does not...   0   0    2

---

您也可以这样做：

df['is'] = df.A.str.count(r'is', flags=re.IGNORECASE)
df['of'] = df.A.str.count(r'of', flags=re.IGNORECASE)
df['she'] = df.A.str.count(r'she', flags=re.IGNORECASE)


                                                   A  of  is  she
0                   Please wait outside of the house   1   0    0
1                    A glittering gem is not enough.   0   1    0
2  The memory we used to share is no longer coher...   0   1    0
3  She only paints with bold colors; she does not...   0   0    2