條件下的Python定義函數

Question

這段代碼在Python中是否正確？

def foo(flag): 
    if flag: 
     def bar(): 
      # Somthing 
    else: 
     def bar(): 
      # Somthing else 
    bar() 

foo(True) 
foo(False) 

如果不是什麼建議的方式來設置一些函數（酒吧）的行爲下？條件？

確定真正的代碼如下

# Building replaceFunc based of ignore_case and use_regexp flags 
if not ignore_case: 
    if not use_regexp: 
     def replaceFunc(string, search, replace): 
      return string.replace(search, replace) 
    else: 
     def replaceFunc(string, search, replace): 
      pattern = re.compile(search) 
      return pattern.sub(replace, string) 
else: 
    if not use_regexp: 
     # There is no standard puthon function for replacing string by ignoring case 
     def replaceFunc(string, search, replace): 
      # implementation from http://stackoverflow.com/questions/919056/case-insensitive-replace 
      return string 
    else: 
     def replaceFunc(string, search, replace): 
      pattern = re.compile(search, re.IGNORECASE) 
      return pattern.sub(replace, string 

Answer 1

這裏有一個合理的方式來實現你想要的：

定義功能的

def bar1(): 
    return 'b1' 

def bar2(): 
    return 'b2' 

def foo(flag): 
    bar = bar2 if flag else bar1 
    return bar() 

print(foo(False)) 
print(foo(True)) 

一個好處bar1()和bar2()外foo()是它們能進行單元測試。