Python代码行太长，需要拆分

Question

如果可能的话，我如何用更少的代码将下面所示代码的结尾部分移到另一行或修改文本，以实现所需的结果。我键入了以下代码：

import pandas as pd
import requests
from bs4 import BeautifulSoup
res = requests.get("http://web.archive.org/web/20070826230746/http://www.bbmf.co.uk/july07.html")
soup = BeautifulSoup(res.content,'lxml')
table = soup.find_all('table')[0]

df = pd.read_html(str(table))
df = df[1]
df = df.rename(columns=df.iloc[0])
df = df.iloc[2:]
df.head(15)

Southport = df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'S') & (df['Hurricane'] == 'H') | (df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'S') | df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'SS')] 
Southport

我试图实现的是显示以下数据：-仅显示，仅显示Dakota Spitfire和Hurricane或Dakota和Spitfire或Dakota和两个Spitfire，如果它们显示在数据表明细表中，则完整代码如下。需要编辑的是从Southport=开始的行：

当我运行代码时，我得到以下回溯错误，我认为这是由于代码行太长：

File "<ipython-input-1-518a9f1c8e98>", line 23
    Southport
            ^
SyntaxError: invalid syntax

我正在运行网络程序Jupyter笔记本里的代码

Answer 1

你有一条长得离谱的线，可以从分裂中获益：

Southport = df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'S') & (df['Hurricane'] == 'H') | (df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'S') | df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'SS')] 

让我们这样做，我们可能会看到错误：

Southport = df[
    df['Location'].str.contains('- Display') & 
    (df['Lancaster'] == '') & 
    (df['Dakota'] == 'D') & 
    (df['Spitfire'] == 'S') & 
    (df['Hurricane'] == 'H') | 
    (df[
        df['Location'].str.contains('- Display') & 
        (df['Lancaster'] == '') & 
        (df['Dakota'] == 'D') & 
        (df['Spitfire'] == 'S') | 
        df[
            df['Location'].str.contains('- Display') & 
            (df['Lancaster'] == '') & 
            (df['Dakota'] == 'D') & 
            (df['Spitfire'] == 'SS')] 

在我看来，括号和括号没有正确对齐——开括号（准确地说，两个开方括号和一个开方括号）比闭括号多。这与为什么您会收到一个SynTaxError是一致的。所以让我们试着解决这个问题：

Southport = df[
        (
            df['Location'].str.contains('- Display') & 
            df['Lancaster'] == '' & 
            df['Dakota'] == 'D' & 
            df['Spitfire'] == 'S' & 
            df['Hurricane'] == 'H'
        )
    ] | df[
        (
            df['Location'].str.contains('- Display') & 
            df['Lancaster'] == '' & 
            df['Dakota'] == 'D' & 
            df['Spitfire'] == 'S'
        )
    ] | df[
        (
            df['Location'].str.contains('- Display') & 
            df['Lancaster'] == '' & 
            df['Dakota'] == 'D' & 
            df['Spitfire'] == 'SS'
        )
    ] 

我想，这与你想写的东西更接近（尽管我不确定你最初到底想写什么）。请注意，将这样的代码分隔成单独的行，这样可以更容易地查看条件是如何对齐的，哪些比较嵌套在哪些其他比较中，等等。

一般的经验法则是，将行长度保持在某个阈值以下（例如，120个字符），您的特定IDE可能会也可能不会对您大喊大叫。有时，由于变量名冗长，所以会出现一个跨越整行的单数表达式，这很好。但总的来说，这种经验法则的存在是为了鼓励这种行为——以一种让阅读代码的人更清楚的方式来拆分行。

Answer 2

我认为这是典型的复制/粘贴错误；您只需删除第一个（第一个（df[）之后的df[（df['Hurricane']='H'）|，以及第二个|之后的df[），那么至少不应该再有任何语法错误。

然而，逻辑过于冗长，因为df['Location']。str.contains（'-Display'）和（df['Lancaster']=''）以及（df['Dakota']='D'）是每个布尔项或分隔布尔项的一部分。

除此之外，df['Location']。str.contains（“-Display”）

Southport = df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'S') | df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & (df['Spitfire'] == 'SS')] 

它可以表示为

Southport = df[(df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D')) & ((df['Spitfire'] == 'S') | (df['Spitfire'] == 'SS'))] 

因为A

如果我没记错的话，像=='S'或=='SS'这样的模式应该用pandas'isin来更好地表达：

Southport = df[df['Location'].str.contains('- Display') & (df['Lancaster'] == '') & (df['Dakota'] == 'D') & df['Spitfire'].isin(['S', 'SS'])]