例如,
test = ["test1", "test2", "test3"]
print ([(i, test) for i, test in enumerate(test)])
>> [(0, 'test1'), (1, 'test2'), (2, 'test3')]
有没有办法改为[('test',0),('test2',1),('test3',2)]?
答案 0 :(得分:1)
使用:
test = ["test", "test2", "test3"]
print ([(test1, i) for i, test1 in enumerate(test)])
我确实修复了您在开始代码中出现的小错字。我将i, test
更改为i, test1
。
我将(i,test1)
切换为(test1,i)
。
答案 1 :(得分:1)
除了明显的genexpr / listcomp包装器:
# Lazy
((x, i) for i, x in enumerate(test))
# Eager
[(x, i) for i, x in enumerate(test)]
你可以使用map
(Py2上的future_builtins.map
)和operator.itemgetter
在C层进行反转以获得额外的速度:
from future_builtins import map # Only on Py2, to get Py3-like generator based map
from operator import itemgetter
# Lazy, wrap in list() if you actually need a list, not just iterating results
map(itemgetter(slice(None, None, -1)), enumerate(test))
# More succinct, equivalent in this case, but not general reversal
map(itemgetter(1, 0), enumerate(test))
答案 2 :(得分:1)
你可以简单地在列表理解语句中切换变量。
test = ["test", "test2", "test3"]
print ([(test,i) for (i,test) in enumerate(test)])
结果:
[('test', 0), ('test2', 1), ('test3', 2)]