collection = ["81851 19AJA01", "68158 17ARB03", "104837 20AAH02",
我有一个像这样扩展的列表,我想根据每个第二部分的前两位数对它们进行排序。我现在真的很急,所以一些帮助会很好。
我尝试过这个并没有用。我不是为了一堂课,我真的很适合帮助
for x in collection:
counter = 0
i=0
for y in len(str(x)):
if (x[i] == '1'):
counter == 1
elif (x[i] == '2'):
counter == 2
elif x[i] == '0' and counter == 2:
counter = 2
elif x[i] == '9' and counter == 1:
counter = 3
elif x[i] == '8' and counter == 1:
counter = 4
elif x[i] == '7' and counter == 1:
counter = 5
i = i + 1
if (counter==2):
freshmen.append(x)
elif (counter==3):
sophomores.append(x)
elif (counter==4):
juniors.append(x)
elif (counter==5):
seniors.append(x)
答案 0 :(得分:3)
使用key
function定义自定义排序规则:
In [1]: collection = ["81851 19AJA01", "68158 17ARB03", "104837 20AAH02"]
In [2]: sorted(collection, key=lambda x: int(x.split()[1][:2]))
Out[2]: ['68158 17ARB03', '81851 19AJA01', '104837 20AAH02']