删除python中的for循环并查找和替换文本

我有两个数据帧。我想在两个数据帧之间进行查找和替换。在

df_find

dataframe的当前标题列中，我想在每一行中搜索

df_replace

数据框中“

keywrod

”列中出现的任何值，如果找到，则用“

keywordlength

”列中的相应值替换它

由于我需要使用

str.replace

函数对该数据帧中的每一行进行迭代，因此我已经能够摆脱

df\u find

数据帧的循环，这是

replace

函数的矢量化形式

在我的例子中，性能很重要，因为两个数据帧都运行在GB中。因此，我想在这里去掉

df_replace

的循环，并使用任何其他有效的方法迭代

df_replace

数据帧的所有行

import pandas as pd
df_find = pd.read_csv("input_find.csv")
df_replace = pd.read_csv("input_replace.csv")

#replace
for i,j in zip(df_replace.keyword,df_replace.keywordLength):
    df_find.current_title=df_find.current_title.str.replace(i,j,case=False)

df_更换 此数据框包含查找和替换所需的数据

keyword       keywordLength
IT Manager    ##10##
Sales Manager ##13##
IT Analyst    ##12##
Store Manager ##13##

df_find是我们需要进行转换的地方

在执行查找和替换代码之前：

current_title
I have been working here as a store manager since after I passed from college
I am sales manager and primarily work in the ASEAN region. My primary rolw is to bring new customers.
I initially joined as a IT analyst and because of my sheer drive and dedication, I was promoted to IT manager position within 3 years

通过上述代码执行查找和替换后

current_title
I have been working here as a ##13## since after I passed from college
I am ##13## and primarily work in the ASEAN region. My primary rolw is to bring new customers.
I initially joined as a ##12## and because of my sheer drive and dedication, I was promoted to ##10## position within 3 years

---

我将永远感激你！谢谢

如果我理解正确，您应该能够对数据集进行相对简单的合并（使用其他几行），并获得所需的结果

没有你的数据集，我只是自己编了一套。下面的代码可能会更优雅一些，但它可以让您在四行中找到需要的位置，最重要的是，没有循环：

设置：

df_find = pd.DataFrame({
            'current_title':['a','a','b','c','b','c','b','a'],
            'other':['this','is','just','a','bunch','of','random','words']
        })

df_replace = pd.DataFrame({'keyword':['a','c'], 'keywordlength':['x','z']})

# This line is to simply re-sort at the end of the code.  Someone with more experience can probably bypass this step.
df_find['idx'] = df_find.index

# Merge together the two data sets based on matching the "current_title" and the "keyword"
dfx = df_find.merge(df_replace, left_on = 'current_title', right_on = 'keyword', how = 'outer').drop('keyword', 1)

# Now, copy the non-null "keywordlength" values to "current_title"
dfx.loc[dfx['keywordlength'].notnull(), 'current_title'] = dfx.loc[dfx['keywordlength'].notnull(), 'keywordlength']

# Clean up by dropping the unnecessary columns and resort based on the first line above.
df_find = dfx.sort_values('idx').drop(['keywordlength','idx'], 1)

  current_title   other
0             x    this
1             x      is
3             b    just
6             z       a
4             b   bunch
7             z      of
5             b  random
2             x   words

代码：

df_find = pd.DataFrame({
            'current_title':['a','a','b','c','b','c','b','a'],
            'other':['this','is','just','a','bunch','of','random','words']
        })

df_replace = pd.DataFrame({'keyword':['a','c'], 'keywordlength':['x','z']})

# This line is to simply re-sort at the end of the code.  Someone with more experience can probably bypass this step.
df_find['idx'] = df_find.index

# Merge together the two data sets based on matching the "current_title" and the "keyword"
dfx = df_find.merge(df_replace, left_on = 'current_title', right_on = 'keyword', how = 'outer').drop('keyword', 1)

# Now, copy the non-null "keywordlength" values to "current_title"
dfx.loc[dfx['keywordlength'].notnull(), 'current_title'] = dfx.loc[dfx['keywordlength'].notnull(), 'keywordlength']

# Clean up by dropping the unnecessary columns and resort based on the first line above.
df_find = dfx.sort_values('idx').drop(['keywordlength','idx'], 1)

  current_title   other
0             x    this
1             x      is
3             b    just
6             z       a
4             b   bunch
7             z      of
5             b  random
2             x   words

输出：

df_find = pd.DataFrame({
            'current_title':['a','a','b','c','b','c','b','a'],
            'other':['this','is','just','a','bunch','of','random','words']
        })

df_replace = pd.DataFrame({'keyword':['a','c'], 'keywordlength':['x','z']})

# This line is to simply re-sort at the end of the code.  Someone with more experience can probably bypass this step.
df_find['idx'] = df_find.index

# Merge together the two data sets based on matching the "current_title" and the "keyword"
dfx = df_find.merge(df_replace, left_on = 'current_title', right_on = 'keyword', how = 'outer').drop('keyword', 1)

# Now, copy the non-null "keywordlength" values to "current_title"
dfx.loc[dfx['keywordlength'].notnull(), 'current_title'] = dfx.loc[dfx['keywordlength'].notnull(), 'keywordlength']

# Clean up by dropping the unnecessary columns and resort based on the first line above.
df_find = dfx.sort_values('idx').drop(['keywordlength','idx'], 1)

  current_title   other
0             x    this
1             x      is
3             b    just
6             z       a
4             b   bunch
7             z      of
5             b  random
2             x   words

---

匹配的值是完全匹配，还是仅子字符串匹配？如果有多个匹配呢？你只是第一场比赛吗？所有的比赛都被替换了。完全匹配。查看正则表达式和

re.sub

。您可以将文件读取为文本，将要替换的内容替换为regex，然后将其作为csv.str打开。replace是re-sub的矢量化实现。它对整个列而不是一行执行操作。这不是我要找的。我添加了一个“之前”和“之后”的示例，以使其更清楚。谢谢好的，这就是为什么我问他们是完全匹配还是子字符串匹配。。。无论如何，这突出了发布数据集的重要性。这是完全匹配的。我们正在用关键字长度替换整个关键字。for循环中的代码就是这样做的。我不知道你说的“完全匹配”是不是指别的意思，但我是按字面意思说的。你能帮我什么忙吗？