[英]Understanding Python List Comprehension equivalent
我有以下代碼:
listOfStrings = ['i_am_exercising', 'python_functional', 'lists_comprehension']
[ "".join([elem.title() for elem in splited]) for splited in [el.split("_")for el in listOfStrings]]
結果是:
['IAmExercising', 'PythonFunctional', 'ListsComprehension']
閱讀文檔 ,我得到了等效的擴展版本,它將第一個表達式放在一個要附加的變量中,第二個表達式放在一個列表中,用for語句進行迭代:
returned = []
for splited in [el.split("_")for el in listOfStrings]:
returned.append("".join([elem.title() for elem in splited]))
但如果我想在沒有任何列表理解的情況下編寫相同的代碼,那么最好的方法是什么? 我嘗試使用以下代碼,效果很好:
returned = []
temp = []
for el in listOfStrings:
temp = []
for splited in el.split("_"):
temp.append(splited.title())
returned.append("".join(temp))
但我沒有完全理解如何做到這一點(將列表理解轉換為等效的完整擴展形式)
你有一個嵌套的列表推導,一個在另一個里面,另外一個用於創建一個拆分元素列表。 您可以將此減少為僅兩個循環:
returned = []
for el in listOfStrings:
tmp = []
for splited in el.split("_"):
tmp.append(splited.title())
returned.append("".join(tmp))
這簡化了對表單的列表理解:
["".join([splited.title() for splited in el.split("_")]) for el in listOfStrings]
您可以輕松地從外部轉換為內部:
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]
變成了
result = []
for bar in baz:
result.append(foo)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.