python修改函数中的列表切片 [英] python Modifying slice of list in function
问题描述
请考虑以下代码:
def func1(a):
a[:] = [x**2 for x in a]
a = range(10)
print a #prints [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
func1(a[:5])
print a #also prints [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
我希望发送列表a
的一部分,并在函数内对其进行更改.我的预期输出是
I wish to send a slice of the list a
and change it inside the function. My expected output is
[0, 1, 4, 9, 16, 5, 6, 7, 8, 9]
哪种方法是惯用的方法?
Which way is the idiomatic way to do so?
谢谢!
推荐答案
如果对列表进行切片,则只能修改一个副本,因此您要执行的操作无法按照您想要的形式进行.
If you slice the list, you modify only a copy, so what you want to do doesn't work in the form you want.
但是您可以将可选的slice
对象传递给func1
,如果不是None
,则使用它来执行切片分配(否则使用[:]
)
But you could pass an optional slice
object to func1
and if it's not None
, use it to perform the slice assignment (else use [:]
)
我将执行以下操作(使用lambda
来避免复制/粘贴公式,并使用生成器表达式来避免创建无用的临时列表:
I would do the following (used a lambda
to avoid copy/paste of the formula and a generator expression to avoid creating a useless temporary list:
def func1(a,the_slice=None):
e = lambda y : (x**2 for x in y)
if the_slice:
a[the_slice] = e(a[the_slice])
else:
a[:] = e(a)
测试:
a = list(range(10))
func1(a)
print(a)
a = list(range(10))
func1(a,slice(5)) # stop at 5
print(a)
a = list(range(10))
func1(a,slice(5,len(a),2)) # start at 5 / stop at the end, stride/step 2
print(a)
结果:
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
[0, 1, 4, 9, 16, 5, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 25, 6, 49, 8, 81]
- 在第一种情况下,列表的总数已更改
- 在第二种情况下,它仅更改了前半部分.
- 在第三种情况下,它更改了下半部分,但是2中有1个值(步幅= 2)
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