通过将它传递给函数来修改python中的列表 [英] Modifying a list in python by passing it to a function

问题描述

我有以下python代码：

  def make_great（l）：
l = [' （魔术师）
打印（魔术师）
   
魔术师= ['Tom'] 
 make_great / pre> 
 
 
 print语句打印原始列表，而不是由 make_great（）函数修改的列表。我试图找出修改python列表的每个元素而不显式访问它的索引，然后重新分配该索引一个新值的方法。现在 magicians 列表不会指向基于Python内存模型的理解所形成的新列表吗？我猜测我没有得到预期的结果，因为这是一个范围问题？但是当我传入 magicians 时，它会替换函数参数中的 l ，所以python解释器实际上会看到 magicians = ['The ... ？
解决方案
 c $ c> l ，你正在重新定义 l ，而不是修改它。使用 l [：] 代替： 
 
 
  def make_great（l）： 
l [：] = ['伟大的魔术师+魔术师] 
  
你也可以返回列表并重新定义魔术师： 
 
 
  def make_great（l ）：
 return ['魔术师的魔术师'] 
 
魔术师= ['Tom'] 
魔术师= make_great（魔术师）
 print（魔术师）
  
在这种情况下，您可以分配魔术师 to  make_great（['Tom']）： 
 
 
 魔术师= make_great（['Tom']）
 print（magicians）
  
 
I have the following python code:
def make_great(l):
    l = ['The great ' + magician for magician in l]

magicians = ['Tom']
make_great(magicians)
print(magicians)
The print statement prints the original list, not the one modified by the make_great() function. I was trying to figure out a way of modifying each element of a python list without explicitly accessing it's index and then reassigning that index a new value. Wouldn't the magicians list now point to the new list formed by the comprehension based on Python's memory model ? I am guessing that I am not getting the desired results because this is a scoping issue ? But when I pass in magicians , it replaces l in the function argument so the python interpreter actually sees magicians = ['The ... ?
解决方案
When you assign it to l, you are redefining l, not modifying it.  Use l[:] instead:
def make_great(l):
    l[:] = ['The great ' + magician for magician in l]
You could also return the list and redefine magicians:
def make_great(l):
    return ['The great ' + magician for magician in l]

magicians = ['Tom']
magicians = make_great(magicians)
print(magicians)
In that case, you could assign magicians to make_great(['Tom']):
magicians = make_great(['Tom'])
print(magicians)


                        
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