在Python中,如何替换字符串列中的字符?

在Python中,如何替换字符串列中的字符?,python,pandas,numpy,dataframe,data-cleaning,Python,Pandas,Numpy,Dataframe,Data Cleaning,这是上面代码的结果 data = pd.read_json('https://raw.githubusercontent.com/OMS1996/DAV-5400/master/Books.json') data= data.transpose() data=data[["Title","Authors","Publisher","Language","Pages"]] data 我想从Author列中删除'['以及']'。我尝试了.replace(,)它不起作用,我也尝试了.str.repl

这是上面代码的结果

data = pd.read_json('https://raw.githubusercontent.com/OMS1996/DAV-5400/master/Books.json')
data= data.transpose()
data=data[["Title","Authors","Publisher","Language","Pages"]]
data
我想从Author列中删除
'['
以及
']'
。我尝试了
.replace(,)
它不起作用,我也尝试了
.str.replace
它不起作用

所有类似的问题都没有帮助

谢谢

试试这个

        Title               Authors                      Publisher                  Language      Pages
Book1   David Copperfield   Charles Dickens              Wordsworth Editions Ltd    English       768
Book2   C how to program    [Havey Deitel, Paul Deitel]  Prentice Hall              English       912
Book3   12 Rules for life   Jordan Peterson              Penguin                                    English       448
试试这个

        Title               Authors                      Publisher                  Language      Pages
Book1   David Copperfield   Charles Dickens              Wordsworth Editions Ltd    English       768
Book2   C how to program    [Havey Deitel, Paul Deitel]  Prentice Hall              English       912
Book3   12 Rules for life   Jordan Peterson              Penguin                                    English       448
您可以尝试:

df = pd.DataFrame({"auth":["Charles Dickens","[Havey Deitel, Paul Deitel]","Jordan Peterson"]})
df["auth"] = df["auth"].str.replace("[","").str.replace("]","")
输出:

data.Authors.apply(lambda x: ', '.join(x) if type(x) == list  else x)
您可以尝试:

df = pd.DataFrame({"auth":["Charles Dickens","[Havey Deitel, Paul Deitel]","Jordan Peterson"]})
df["auth"] = df["auth"].str.replace("[","").str.replace("]","")
输出:

data.Authors.apply(lambda x: ', '.join(x) if type(x) == list  else x)
  • 试试这个-

                       Title                    Authors                Publisher Language Pages
    Book1  David Copperfield            Charles Dickens  Wordsworth Editions Ltd  English   768
    Book2   C how to program  Havey Deitel, Paul Deitel            Prentice Hall  English   912
    Book3  12 Rules for life            Jordan Peterson                  Penguin  English   448
    
  • 试试这个-

                       Title                    Authors                Publisher Language Pages
    Book1  David Copperfield            Charles Dickens  Wordsworth Editions Ltd  English   768
    Book2   C how to program  Havey Deitel, Paul Deitel            Prentice Hall  English   912
    Book3  12 Rules for life            Jordan Peterson                  Penguin  English   448
    

  • 它是
    列表
    。只需运行此命令即可查看输出

    data['Authors'] = data['Authors'].apply(lambda x: ",".join(x) if isinstance(x, list) else x)
    data['Authors'] = data['Authors'].str.replace(",", " ")
    
    要将它们转换为字符串和条带
    []
    ,只需映射
    repr
    并将
    []'
    替换为
    '


    它是
    列表
    。只需运行此命令即可查看输出

    data['Authors'] = data['Authors'].apply(lambda x: ",".join(x) if isinstance(x, list) else x)
    data['Authors'] = data['Authors'].str.replace(",", " ")
    
    要将它们转换为字符串和条带
    []
    ,只需映射
    repr
    并将
    []'
    替换为
    '

    它不起作用,请详细说明。你到底做了什么,它抛出了一个错误,结果是什么样的,等等。?数据帧输出的格式设置已关闭。它不起作用。请详细说明。你到底做了什么,它抛出了一个错误,结果是什么样的,等等。?数据帧输出的格式设置已关闭。