是这样的
def mymethod():
return [[1,2,3,4],
[1,2,3,4],
[1,2,3,4],
[1,2,3,4]]
mylist = mymethod()
for _, thing, _, _ in mylist:
print thing
# this bit is meant to be outside the for loop,
# I mean it to represent the last value thing was in the for
if thing:
print thing
我想要做的是避免虚拟变量，有没有比
更聪明的方法for thing in mylist:
print thing[1]
因为那时我必须在我需要的任何其他时间使用thing[1]，而不是将它分配给一个新的变量然后事情变得一团糟。
new to to python如果我遗漏了一些明显的东西，那就很抱歉
答案 0 :(得分：7)
你可以破解生成器表达式
def mymethod():
return [[1,2,3,4],
[1,2,3,4],
[1,2,3,4],
[1,2,3,4]]
mylist = mymethod()
for thing in (i[1] for i in mylist):
print thing
# this bit is meant to be outside the for loop,
# I mean it to represent the last value thing was in the for
if thing:
print thing
答案 1 :(得分：3)
如果你想得到一个数组的第二列，你可以使用列表推导，如下所示：
a = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9,10,11,12 ],
[13,14,15,16 ] ]
second_column = [ row[1] for row in a ]
# you get [2, 6, 10, 14]
你可以把它包装成一个函数：
def get_column ( array, column_number ):
try:
return [row[column_number] for row in array]
except IndexError:
print ("Not enough columns!")
raise # Raise the exception again as we haven't dealt with the issue.
fourth_column = get_column(a,3)
# you get [4, 8, 12, 16]
tenth_column = get_column(a,9)
# You requested the tenth column of a 4-column array, so you get the "not enough columns!" message.
尽管如此，如果您正在使用矩形数组，您希望使用numpy数组，而不是数字列表列表。
或者，根据Lattyware的暗示请求，生成器版本：
def column_iterator ( array, column_number ):
try:
for row in array:
yield row[column_number]
except IndexError:
print ("Not enough columns!")
raise
用法就像普通列表一样：
>>> for item in column_iterator(a,1):
... print(item)
...
2
6
10
14
>>>
发电机性质很明显：
>>> b = column_iterator(a,1)
>>> b.next()
2
>>> b.next()
6
>>> b.next()
10
>>> b.next()
14
答案 2 :(得分：2)
当然，itertools.chain和slicing何时会提供帮助？
for thing in itertools.islice(itertools.chain(*mylist),1,None,len(mylist)):
print(thing)
Numpy对列切片也很有帮助。这是numpy中的另一个例子
for thing in numpy.array(mylist)[:,1]:
print(thing)
答案 3 :(得分：1)
虽然我喜欢Dikei's answer的清晰度和简洁性，但我仍然认为一个好的选择就是：
for sublist in mylist:
item = sublist[1]
...
do_stuff(item)
...
do_other_stuff(item)
...
很明显，可以扩展到更容易做，而且可能是最快的。
以下是一些快速测试 - 由于在循环中什么都不做，我不确定它们会有多准确，但是他们可能会提出一个想法：
python -m timeit -s "mylist = [range(1,8) for _ in range(1,8)]" 'for thing in mylist:' ' item=thing[1]' ' pass'
1000000 loops, best of 3: 1.25 usec per loop
python -m timeit -s "mylist = [range(1,8) for _ in range(1,8)]" 'for thing in (i[1] for i in mylist):' ' pass'
100000 loops, best of 3: 2.37 usec per loop
python -m timeit -s "mylist = [range(1,8) for _ in range(1,8)]" 'for thing in itertools.islice(itertools.chain(*mylist),1,None,len(mylist)):' ' pass'
1000000 loops, best of 3: 2.21 usec per loop
python -m timeit -s "import numpy" -s "mylist = numpy.array([range(1,8) for _ in range(1,8)])" 'for thing in mylist[:,1]:' ' pass'
1000000 loops, best of 3: 1.7 usec per loop
python -m timeit -s "import numpy" -s "mylist = [range(1,8) for _ in range(1,8)]" 'for thing in numpy.array(mylist)[:,1]:' ' pass'
10000 loops, best of 3: 63.8 usec per loop
请注意，numpy如果生成一次就很快，但是对于单个操作按需生成非常慢。
在大型名单上：
python -m timeit -s "mylist = [range(1,100) for _ in range(1,100)]" 'for thing in mylist:' ' item=thing[1]' ' pass'
100000 loops, best of 3: 16.3 usec per loop
python -m timeit -s "mylist = [range(1,100) for _ in range(1,100)]" 'for thing in (i[1] for i in mylist):' ' pass'
10000 loops, best of 3: 27 usec per loop
python -m timeit -s "mylist = [range(1,100) for _ in range(1,100)]" 'for thing in itertools.islice(itertools.chain(*mylist),1,None,len(mylist)):' ' pass'
10000 loops, best of 3: 101 usec per loop
python -m timeit -s "import numpy" -s "mylist = numpy.array([range(1,100) for _ in range(1,100)])" 'for thing in mylist[:,1]:' ' pass'
100000 loops, best of 3: 8.47 usec per loop
python -m timeit -s "import numpy" -s "mylist = [range(1,100) for _ in range(1,100)]" 'for thing in numpy.array(mylist)[:,1]:' ' pass'
100 loops, best of 3: 3.82 msec per loop
请记住，除非真的需要速度，否则速度应始终位于可读性的第二位。
答案 4 :(得分：1)
方法itemgetter()可用于解决此问题：
from operator import itemgetter
def mymethod():
return [[1,2,3,4],
[1,2,3,4],
[1,2,3,4],
[1,2,3,4]]
mylist = mymethod()
row = map(itemgetter(2), mylist)
print("row %s" % row)
thing = row[-1]
# this bit is meant to be outside the for loop,
# I mean it to represent the last value thing was in the for
if thing:
print thing
输出结果为：
row [3, 3, 3, 3]
3