a = [范围（4）中n的[λt：t ** n]] [英] a=[ lambda t: t**n for n in range(4) ]

问题描述

我正在考虑以下内容;

I was thinking about something like the following;

a = [t ** n for n in范围（4）]
回溯（最近一次调用最后一次）：

文件"< stdin>"，第1行，在？

NameError：名称''t''未定义


或

a = [lambda t：t ** n n为范围（4）]
t = 2
a
[< function< lambda>在0x403dcc6c>，< function< lambda>在0x403dcca4>，

< function< lambda>在0x403dccdc>，< function< lambda>在0x403dcd14>] t = 3
a
[< function< lambda>在0x403dcc6c>，< function< lambda>在0x403dcca4>，

< function< lambda>在0x403dccdc>，< function< lambda>在0x403dcd14>]

a=[ t**n for n in range(4) ] Traceback (most recent call last):
File "<stdin>", line 1, in ?
NameError: name ''t'' is not defined
or
a=[ lambda t: t**n for n in range(4) ]
t=2
a [<function <lambda> at 0x403dcc6c>, <function <lambda> at 0x403dcca4>,
<function <lambda> at 0x403dccdc>, <function <lambda> at 0x403dcd14>] t=3
a [<function <lambda> at 0x403dcc6c>, <function <lambda> at 0x403dcca4>,
<function <lambda> at 0x403dccdc>, <function <lambda> at 0x403dcd14>]

有可能吗？你能给我一些建议吗？

is something like that possible? Will you give me advice about that?

推荐答案

我************* @ gmail.com 写道：

我在考虑以下内容;

I was thinking about something like the following;

a = [t ** n for n in range（4）]
Traceback（最近一次调用最后一次） ）：
文件"< stdin>"，第1行，在？
NameError：名称''t''未定义

或
< br => a = [lambda t：t ** n为范围内的n（4）]
t = 2
a


也许你的意思是：a = lambda t：[t ** n代表范围内的n（4）]
a（2）
[1,2,4,8]


但是写得更好：def a（t）：
...返回[t ** n代表范围内的n（4）]

...
a （2）
[1,2,4,8]

a=[ t**n for n in range(4) ]
Traceback (most recent call last):
File "<stdin>", line 1, in ?
NameError: name ''t'' is not defined
or

a=[ lambda t: t**n for n in range(4) ]
t=2
a
Perhaps you mean: a = lambda t: [t**n for n in range(4)]
a(2) [1, 2, 4, 8]
but that is better written as: def a(t): ... return [t**n for n in range(4)]
...
a(2) [1, 2, 4, 8]

Michael

Michael

请发布你的tr的描述ying to

完成而不是要求我们采取你可能的

解决方案并试图找出问题所在。


你可以这样做：


def foo（t，r）：

返回[t ** n表示范围内的r（r）]


a = foo（2,4）

a

[1,2,4,8]


a = foo（3,6）

a

[1,3,9,27,82,243]


但我真的不知道你想要什么。


拉里
我************* @ gmail.com 写道：

Please post a description of what you are trying to
accomplish instead of asking us to take your possible
solution and try to figure out what the problem was.

You can do:

def foo(t, r):
return [t**n for n in range(r)]

a=foo(2, 4)
a
[1, 2, 4, 8]

a=foo(3,6)
a
[1, 3, 9, 27, 81, 243]

but I don''t really know what you want.

Larry
me*************@gmail.com wrote:

我正在考虑以下内容;

I was thinking about something like the following;

a = [t ** n for n in range（4）]
Traceback（最近一次呼叫最后一次）：
文件"< stdin>"，第1行，在？
NameError：名称''t''未定义

或

a = [lambda t：t ** n，n为范围（4）]
t = 2
a
[< function< lambda >在0x403dcc6c>，< function< lambda>在0x403dcca4>，
< function< lambda>在0x403dccdc>，< function< lambda>在0x403dcd14>]
t = 3
a

a=[ t**n for n in range(4) ]
Traceback (most recent call last):
File "<stdin>", line 1, in ?
NameError: name ''t'' is not defined
or

a=[ lambda t: t**n for n in range(4) ]
t=2
a
[<function <lambda> at 0x403dcc6c>, <function <lambda> at 0x403dcca4>,
<function <lambda> at 0x403dccdc>, <function <lambda> at 0x403dcd14>]
t=3
a

[< function< lambda>在0x403dcc6c>，< function< lambda>在0x403dcca4>，
< function< lambda>在0x403dccdc>，< function< lambda>在0x403dcd14>]

有可能吗？你能给我一些建议吗？

[<function <lambda> at 0x403dcc6c>, <function <lambda> at 0x403dcca4>,
<function <lambda> at 0x403dccdc>, <function <lambda> at 0x403dcd14>]
is something like that possible? Will you give me advice about that?

Thanx你的回复。


我正在寻找函数数组。

类似a = [sin（x），cos（x）]

Thanx for your replies.

I''m looking for array of functions.
Something like a=[ sin(x) , cos(x) ]

x = 0.0
a
[0,1] x = 1.0
a
....


当然它可以通过def cratearray（x）制作：

x=0.0
a [0, 1] x=1.0
a ....

of course it can be made by def cratearray(x):

.... ~~~~

。 ...返回

a = createarray（1.0）


但这不是我要求的。自动化的东西。

.... ~~~~
.... return a
a=createarray(1.0)

but this isn''t what i am asking for. something automized.

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