如何根據位置值拆分列表元素

Question

我正在嘗試從一組數字中分割年份、月份和控制數字。

s = ['201911007', '201203008']
my expected output
year=['2019', '2012']
month=['11','01']
controlnum=['007','008']

Answer 1

您可以使用列表理解來生成每個部分的元組，然后 zip 將元組分解為單獨的列表：

s = ['201911007', '201203008']

year,month,controlNum = map(list,zip(*( (v[:4],v[4:6],v[6:]) for v in s )))

output：

print(year)
print(month)
print(controlNum)

['2019', '2012']
['11', '03']
['007', '008']

Answer 2

s = ['201911007', '201203008']
year = []
month = []
controlnum = []
for number in s:
    year.append(number[:4])
    month.append(number[4:6])
    controlnum.append(number[6:])

print(year)
print(month)
print(controlnum)

Answer 3

嘗試使用 function

s = ['201911007', '201203008']
def info(s):
     return [i[:4] for i in s], [i[4:s] for i in a], [i[6:] for i in s]
year, month, controlnum = defo(s)
print (year)