Question

如何将数组中的分隔元素拆分为其中包含分隔元素的新元素。

想法是要更改此

[ "Tiger Nixon, System Architect, Edinburgh, 5421, 2011/04/25", "Garrett Winters, Accountant, Tokyo, 422, 2011/07/25" ]

对此

[
  [ "Tiger Nixon", "System Architect", "Edinburgh", "5421", "2011/04/25" ],
  [ "Garrett Winters", "Accountant", "Tokyo", "8422", "2011/07/25"]
]

我尝试了此代码，但没有成功。我将my_string设置为顶部数组。

my_list = my_string.split(",")

Answer 1

尝试一下：

# initial list
mystring = [ "Tiger Nixon, System Architect, Edinburgh, 5421, 2011/04/25", "Garrett Winters, Accountant, Tokyo, 422, 2011/07/25" ]
# empty list to store new values 
array = []
# loop through the list and split each value 
for i in mystring:
  array.append(i.split(",")) # splits into list and appends it a new list 
print(array) # prints the resultant array

您还可以使用以下一种划线员列表理解方法。

mystring = [string.split(",") for string in mystring]

输出：

[['Tiger Nixon', ' System Architect', ' Edinburgh', ' 5421', ' 2011/04/25'], ['Garrett Winters', ' Accountant', ' Tokyo', ' 422', ' 2011/07/25']]

查看实际使用的代码here。

Answer 2

[i.split(',') for i in list_of_words]

输出：

[['Tiger Nixon', ' System Architect', ' Edinburgh', ' 5421', ' 2011/04/25'], ['Garrett Winters', ' Accountant', ' Tokyo', ' 422', ' 2011/07/25']]

我认为这有帮助！

Answer 3

将list comprehension与split()一起使用

l = [ "Tiger Nixon, System Architect, Edinburgh, 5421, 2011/04/25", "Garrett Winters, Accountant, Tokyo, 422, 2011/07/25" ]

print([i.split(",") for i in l])

输出：

[['Tiger Nixon', ' System Architect', ' Edinburgh', ' 5421', ' 2011/04/25'],
 ['Garrett Winters', ' Accountant', ' Tokyo', ' 422', ' 2011/07/25']]

将元素数组更改为多个元素数组

3 个答案: