[英]Understanding Python List Comprehension equivalent
我有以下代码:
listOfStrings = ['i_am_exercising', 'python_functional', 'lists_comprehension']
[ "".join([elem.title() for elem in splited]) for splited in [el.split("_")for el in listOfStrings]]
结果是:
['IAmExercising', 'PythonFunctional', 'ListsComprehension']
阅读文档 ,我得到了等效的扩展版本,它将第一个表达式放在一个要附加的变量中,第二个表达式放在一个列表中,用for语句进行迭代:
returned = []
for splited in [el.split("_")for el in listOfStrings]:
returned.append("".join([elem.title() for elem in splited]))
但如果我想在没有任何列表理解的情况下编写相同的代码,那么最好的方法是什么? 我尝试使用以下代码,效果很好:
returned = []
temp = []
for el in listOfStrings:
temp = []
for splited in el.split("_"):
temp.append(splited.title())
returned.append("".join(temp))
但我没有完全理解如何做到这一点(将列表理解转换为等效的完整扩展形式)
你有一个嵌套的列表推导,一个在另一个里面,另外一个用于创建一个拆分元素列表。 您可以将此减少为仅两个循环:
returned = []
for el in listOfStrings:
tmp = []
for splited in el.split("_"):
tmp.append(splited.title())
returned.append("".join(tmp))
这简化了对表单的列表理解:
["".join([splited.title() for splited in el.split("_")]) for el in listOfStrings]
您可以轻松地从外部转换为内部:
listOfStrings = ['i_am_exercising', 'python_functional', 'lists_comprehension']
result = [ "".join([elem.title() for elem in split]) for split in [el.split("_")for el in listOfStrings]]
print result
result = []
for split in [el.split("_") for el in listOfStrings]:
result.append("".join([elem.title() for elem in split]))
print result
result = []
temp1 = []
for el in listOfStrings:
temp1.append(el.split("_"))
for split in temp1:
result.append("".join([elem.title() for elem in split]))
print result
result = []
temp1 = []
for el in listOfStrings:
temp1.append(el.split("_"))
for split in temp1:
temp2 = []
for elem in split:
temp2.append(elem.title())
result.append("".join(temp2))
print result
基本上你只需遵循以下方案:
result = [foo for bar in baz]
变成了
result = []
for bar in baz:
result.append(foo)
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