python lambda表达式中可以有多个语句吗?

问题:python lambda表达式中可以有多个语句吗?

我是python新手,试图实现以下目标:

我有一个清单清单:

lst = [[567,345,234],[253,465,756, 2345],[333,777,111, 555]]

我希望将lst映射到另一个列表,该列表仅包含每个子列表中第二小的数字。因此结果应为:

[345, 465, 333]

例如,如果我只是对最小的数量感兴趣,我可以这样做:

map(lambda x: min(x),lst)

我希望我可以这样做:

map(lambda x: sort(x)[1],lst)

但是排序不会连锁。(返回无)

不允许这样的事情:

map(lambda x: sort(x); x[1],lst) #hence the multiple statement question

有没有一种方法可以在python中使用map来执行此操作,而无需定义命名函数?(例如,使用Ruby中的匿名块很容易)

I am a python newbie trying to achieve the following:

I have a list of lists:

lst = [[567,345,234],[253,465,756, 2345],[333,777,111, 555]]

I want map lst into another list containing only the second smallest number from each sublist. So the result should be:

[345, 465, 333]

For example if I were just interested in the smallest number, I could do:

map(lambda x: min(x),lst)

I wish I could do this:

map(lambda x: sort(x)[1],lst)

but sort does not chain. (returns None)

neither is something like this allowed:

map(lambda x: sort(x); x[1],lst) #hence the multiple statement question

Is there a way to do this with map in python but without defining a named function? (it is easy with anonymous blocks in ruby, for example)


回答 0

从您的具体问题到更一般的问题,我可以在这里提供几种不同的答案。因此,从最具体到最一般:

问:能否在lambda中放入多个语句?

:不但是您实际上不需要使用lambda。您可以将语句放在a中def。即:

def second_lowest(l):
    l.sort()
    return l[1]

map(second_lowest, lst)

问:您可以通过对列表进行排序从lambda中获取第二低的项目吗?

答:是的。正如alex的答案所指出的那样,sorted()是sort的一种版本,它创建一个新列表,而不是就地排序,并且可以被链接。请注意,这可能是您应该使用的-地图在原始列表上产生副作用是不好的做法。

问:如何从一系列列表中的每个列表中获取第二低的项目?

sorted(l)[1]:实际上并不是最好的方法。它具有O(N log(N))复杂度,同时存在O(n)解决方案。可以在heapq模块中找到。

>>> import  heapq
>>> l = [5,2,6,8,3,5]
>>> heapq.nsmallest(l, 2)
[2, 3]

因此,只需使用:

map(lambda x: heapq.nsmallest(x,2)[1],  list_of_lists)

通常也认为使用列表理解更为清晰,这完全避免了lambda:

[heapq.nsmallest(x,2)[1] for x in list_of_lists]

There are several different answers I can give here, from your specific question to more general concerns. So from most specific to most general:

Q. Can you put multiple statements in a lambda?

A. No. But you don’t actually need to use a lambda. You can put the statements in a def instead. i.e.:

def second_lowest(l):
    l.sort()
    return l[1]

map(second_lowest, lst)

Q. Can you get the second lowest item from a lambda by sorting the list?

A. Yes. As alex’s answer points out, sorted() is a version of sort that creates a new list, rather than sorting in-place, and can be chained. Note that this is probably what you should be using – it’s bad practice for your map to have side effects on the original list.

Q. How should I get the second lowest item from each list in a sequence of lists?

A. sorted(l)[1] is not actually the best way for this. It has O(N log(N)) complexity, while an O(n) solution exists. This can be found in the heapq module.

>>> import  heapq
>>> l = [5,2,6,8,3,5]
>>> heapq.nsmallest(l, 2)
[2, 3]

So just use:

map(lambda x: heapq.nsmallest(x,2)[1],  list_of_lists)

It’s also usually considered clearer to use a list comprehension, which avoids the lambda altogether:

[heapq.nsmallest(x,2)[1] for x in list_of_lists]

回答 1

将语句放在列表中可能会模拟多个语句:

例如:

lambda x: [f1(x), f2(x), f3(x), x+1]

Putting the expressions in a list may simulate multiple expressions:

E.g.:

lambda x: [f1(x), f2(x), f3(x), x+1]

This will not work with statements.


回答 2

时光旅行者在这里。如果通常要在一个lambda中包含多个语句,则可以将其他lambda作为该lambda的参数传递。

(lambda x, f: list((y[1] for y in f(x))))(lst, lambda x: (sorted(y) for y in x))

您实际上不能有多个语句,但是可以通过将lambda传递给lambda来模拟该语句。

编辑:时间旅行者回来了!您还可以将布尔表达式的行为(记住短路规则和真实性)用于链式操作。使用三元运算符可为您提供更多功能。同样,您不能有多个语句,但是您当然可以有许多函数调用。此示例对一堆数据进行了任意处理,但是,它表明您可以做一些有趣的事情。print语句是返回函数的示例None.sort()方法也是如此),但它们也有助于显示函数的作用lambda

>>> (lambda x: print(x) or x+1)(10)
10
11
>>> f = (lambda x: x[::2] if print(x) or x.sort() else print(enumerate(x[::-1]) if print(x) else filter(lambda (i, y): print((i, y)) or (i % 3 and y % 2), enumerate(x[::-1]))))
>>> from random import shuffle
>>> l = list(range(100))
>>> shuffle(l)
>>> f(l)
[84, 58, 7, 99, 17, 14, 60, 35, 12, 56, 26, 48, 55, 40, 28, 52, 31, 39, 43, 96, 64, 63, 54, 37, 79, 25, 46, 72, 10, 59, 24, 68, 23, 13, 34, 41, 94, 29, 62, 2, 50, 32, 11, 97, 98, 3, 70, 93, 1, 36, 87, 47, 20, 73, 45, 0, 65, 57, 6, 76, 16, 85, 95, 61, 4, 77, 21, 81, 82, 30, 53, 51, 42, 67, 74, 8, 15, 83, 5, 9, 78, 66, 44, 27, 19, 91, 90, 18, 49, 86, 22, 75, 71, 88, 92, 33, 89, 69, 80, 38]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
(0, 99)
(1, 98)
(2, 97)
(3, 96)
(4, 95)
(5, 94)
(6, 93)
(7, 92)
(8, 91)
(9, 90)
(10, 89)
(11, 88)
(12, 87)
(13, 86)
(14, 85)
(15, 84)
(16, 83)
(17, 82)
(18, 81)
(19, 80)
(20, 79)
(21, 78)
(22, 77)
(23, 76)
(24, 75)
(25, 74)
(26, 73)
(27, 72)
(28, 71)
(29, 70)
(30, 69)
(31, 68)
(32, 67)
(33, 66)
(34, 65)
(35, 64)
(36, 63)
(37, 62)
(38, 61)
(39, 60)
(40, 59)
(41, 58)
(42, 57)
(43, 56)
(44, 55)
(45, 54)
(46, 53)
(47, 52)
(48, 51)
(49, 50)
(50, 49)
(51, 48)
(52, 47)
(53, 46)
(54, 45)
(55, 44)
(56, 43)
(57, 42)
(58, 41)
(59, 40)
(60, 39)
(61, 38)
(62, 37)
(63, 36)
(64, 35)
(65, 34)
(66, 33)
(67, 32)
(68, 31)
(69, 30)
(70, 29)
(71, 28)
(72, 27)
(73, 26)
(74, 25)
(75, 24)
(76, 23)
(77, 22)
(78, 21)
(79, 20)
(80, 19)
(81, 18)
(82, 17)
(83, 16)
(84, 15)
(85, 14)
(86, 13)
(87, 12)
(88, 11)
(89, 10)
(90, 9)
(91, 8)
(92, 7)
(93, 6)
(94, 5)
(95, 4)
(96, 3)
(97, 2)
(98, 1)
(99, 0)
[(2, 97), (4, 95), (8, 91), (10, 89), (14, 85), (16, 83), (20, 79), (22, 77), (26, 73), (28, 71), (32, 67), (34, 65), (38, 61), (40, 59), (44, 55), (46, 53), (50, 49), (52, 47), (56, 43), (58, 41), (62, 37), (64, 35), (68, 31), (70, 29), (74, 25), (76, 23), (80, 19), (82, 17), (86, 13), (88, 11), (92, 7), (94, 5), (98, 1)]

Time traveler here. If you generally want to have multiple statements within a lambda, you can pass other lambdas as arguments to that lambda.

(lambda x, f: list((y[1] for y in f(x))))(lst, lambda x: (sorted(y) for y in x))

You can’t actually have multiple statements, but you can simulate that by passing lambdas to lambdas.

Edit: The time traveler returns! You can also abuse the behavior of boolean expressions (keeping in mind short-circuiting rules and truthiness) to chain operations. Using the ternary operator gives you even more power. Again, you can’t have multiple statements, but you can of course have many function calls. This example does some arbitrary junk with a bunch of data, but, it shows that you can do some funny stuff. The print statements are examples of functions which return None (as does the .sort() method) but they also help show what the lambda is doing.

>>> (lambda x: print(x) or x+1)(10)
10
11
>>> f = (lambda x: x[::2] if print(x) or x.sort() else print(enumerate(x[::-1]) if print(x) else filter(lambda (i, y): print((i, y)) or (i % 3 and y % 2), enumerate(x[::-1]))))
>>> from random import shuffle
>>> l = list(range(100))
>>> shuffle(l)
>>> f(l)
[84, 58, 7, 99, 17, 14, 60, 35, 12, 56, 26, 48, 55, 40, 28, 52, 31, 39, 43, 96, 64, 63, 54, 37, 79, 25, 46, 72, 10, 59, 24, 68, 23, 13, 34, 41, 94, 29, 62, 2, 50, 32, 11, 97, 98, 3, 70, 93, 1, 36, 87, 47, 20, 73, 45, 0, 65, 57, 6, 76, 16, 85, 95, 61, 4, 77, 21, 81, 82, 30, 53, 51, 42, 67, 74, 8, 15, 83, 5, 9, 78, 66, 44, 27, 19, 91, 90, 18, 49, 86, 22, 75, 71, 88, 92, 33, 89, 69, 80, 38]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
(0, 99)
(1, 98)
(2, 97)
(3, 96)
(4, 95)
(5, 94)
(6, 93)
(7, 92)
(8, 91)
(9, 90)
(10, 89)
(11, 88)
(12, 87)
(13, 86)
(14, 85)
(15, 84)
(16, 83)
(17, 82)
(18, 81)
(19, 80)
(20, 79)
(21, 78)
(22, 77)
(23, 76)
(24, 75)
(25, 74)
(26, 73)
(27, 72)
(28, 71)
(29, 70)
(30, 69)
(31, 68)
(32, 67)
(33, 66)
(34, 65)
(35, 64)
(36, 63)
(37, 62)
(38, 61)
(39, 60)
(40, 59)
(41, 58)
(42, 57)
(43, 56)
(44, 55)
(45, 54)
(46, 53)
(47, 52)
(48, 51)
(49, 50)
(50, 49)
(51, 48)
(52, 47)
(53, 46)
(54, 45)
(55, 44)
(56, 43)
(57, 42)
(58, 41)
(59, 40)
(60, 39)
(61, 38)
(62, 37)
(63, 36)
(64, 35)
(65, 34)
(66, 33)
(67, 32)
(68, 31)
(69, 30)
(70, 29)
(71, 28)
(72, 27)
(73, 26)
(74, 25)
(75, 24)
(76, 23)
(77, 22)
(78, 21)
(79, 20)
(80, 19)
(81, 18)
(82, 17)
(83, 16)
(84, 15)
(85, 14)
(86, 13)
(87, 12)
(88, 11)
(89, 10)
(90, 9)
(91, 8)
(92, 7)
(93, 6)
(94, 5)
(95, 4)
(96, 3)
(97, 2)
(98, 1)
(99, 0)
[(2, 97), (4, 95), (8, 91), (10, 89), (14, 85), (16, 83), (20, 79), (22, 77), (26, 73), (28, 71), (32, 67), (34, 65), (38, 61), (40, 59), (44, 55), (46, 53), (50, 49), (52, 47), (56, 43), (58, 41), (62, 37), (64, 35), (68, 31), (70, 29), (74, 25), (76, 23), (80, 19), (82, 17), (86, 13), (88, 11), (92, 7), (94, 5), (98, 1)]

回答 3

使用排序函数,如下所示:

map(lambda x: sorted(x)[1],lst)

Use sorted function, like this:

map(lambda x: sorted(x)[1],lst)

回答 4

实际上,您可以在python中的lambda表达式中包含多个语句。这并不简单,但在您的示例中,以下工作原理:

map(lambda x: x.sort() or x[1],lst)

您必须确保每个语句不返回任何内容,或者是否将其包装在(..和False)中。结果是最后一次评估返回的结果。

例:

>>> f = (lambda : (print(1) and False) or (print(2) and False) or (print(3) and False))
>>> f()
1
2
3

You can in fact have multiple statements in a lambda expression in python. It is not entirely trivial but in your example, the following works:

map(lambda x: x.sort() or x[1],lst)

You have to make sure that each statement does not return anything or if it does wrap it in (.. and False). The result is what is returned by the last evaluation.

Example:

>>> f = (lambda : (print(1) and False) or (print(2) and False) or (print(3) and False))
>>> f()
1
2
3

回答 5

从此处使用begin():http : //www.reddit.com/r/Python/comments/hms4z/ask_pyreddit_if_you_were_making_your_own/c1wycci

Python 3.2 (r32:88445, Mar 25 2011, 19:28:28) 
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> lst = [[567,345,234],[253,465,756, 2345],[333,777,111, 555]]
>>> begin = lambda *args: args[-1]
>>> list(map(lambda x: begin(x.sort(), x[1]), lst))
[345, 465, 333]

Using begin() from here: http://www.reddit.com/r/Python/comments/hms4z/ask_pyreddit_if_you_were_making_your_own/c1wycci

Python 3.2 (r32:88445, Mar 25 2011, 19:28:28) 
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> lst = [[567,345,234],[253,465,756, 2345],[333,777,111, 555]]
>>> begin = lambda *args: args[-1]
>>> list(map(lambda x: begin(x.sort(), x[1]), lst))
[345, 465, 333]

回答 6

在Python中将多个语句组合成单个语句的Hacky方法是使用“ and”关键字作为短路运算符。然后,您可以将此单个语句直接用作lambda表达式的一部分。

这类似于在shell语言(例如bash)中将“ &&”用作短路运算符。

另请注意:您始终可以通过包装函数来修复函数语句以返回真值。

例:

def p2(*args):
    print(*args)
    return 1 # a true value

junky = lambda x, y: p2('hi') and p2('there') and p2(x) and p2(y)

junky("a", "b")

再三考虑,最好使用“或”代替“与”,因为许多函数成功返回“ 0”或“无”。然后,您可以在上面的示例中摆脱包装函数:

junky = lambda x, y: print('hi') or print('there') or print(x) or print(y)

junky("a", "b")

“ and”运算将对表达式进行求值,直到达到第一个零返回值为止。之后短路。1和1以及0和1求值:1和1和0并丢弃1

“或”运算将对表达式进行求值,直到达到第一个非零返回值为止。之后短路。

0或0或1或0计算0或0或1,并丢弃0

A Hacky way to combine multiple statements into a single statement in python is to use the “and” keyword as a short-circuit operator. Then you can use this single statement directly as part of the lambda expression.

This is similar to using “&&” as the short-circuit operator in shell languages such as bash.

Also note: You can always fix a function statement to return a true value by wrapping the function.

Example:

def p2(*args):
    print(*args)
    return 1 # a true value

junky = lambda x, y: p2('hi') and p2('there') and p2(x) and p2(y)

junky("a", "b")

On second thought, its probably better to use ‘or’ instead of ‘and’ since many functions return ‘0’ or None on success. Then you can get rid of the wrapper function in the above example:

junky = lambda x, y: print('hi') or print('there') or print(x) or print(y)

junky("a", "b")

‘and’ operate will evaluate the expressions until it gets to the first zero return value. after which it short-circuits. 1 and 1 and 0 and 1 evaluates: 1 and 1 and 0, and drops 1

‘or’ operate will evaluate the expressions until it gets to the first non-zero return value. after which it short-circuits.

0 or 0 or 1 or 0 evaluates 0 or 0 or 1, and drops 0


回答 7

或者,如果您想避免使用lambda并使用生成器而不是列表:

(第一个col的sorted(col)[1])

Or if you want to avoid lambda and have a generator instead of a list:

(sorted(col)[1] for col in lst)

回答 8

让我向您介绍一个光荣却可怕的技巧:

import types

def _obj():
  return lambda: None

def LET(bindings, body, env=None):
  '''Introduce local bindings.
  ex: LET(('a', 1,
           'b', 2),
          lambda o: [o.a, o.b])
  gives: [1, 2]

  Bindings down the chain can depend on
  the ones above them through a lambda.
  ex: LET(('a', 1,
           'b', lambda o: o.a + 1),
          lambda o: o.b)
  gives: 2
  '''
  if len(bindings) == 0:
    return body(env)

  env = env or _obj()
  k, v = bindings[:2]
  if isinstance(v, types.FunctionType):
    v = v(env)

  setattr(env, k, v)
  return LET(bindings[2:], body, env)

您现在可以按以下LET方式使用此表单:

map(lambda x: LET(('_', x.sort()),
                  lambda _: x[1]),
    lst)

这使: [345, 465, 333]

Let me present to you a glorious but terrifying hack:

import types

def _obj():
  return lambda: None

def LET(bindings, body, env=None):
  '''Introduce local bindings.
  ex: LET(('a', 1,
           'b', 2),
          lambda o: [o.a, o.b])
  gives: [1, 2]

  Bindings down the chain can depend on
  the ones above them through a lambda.
  ex: LET(('a', 1,
           'b', lambda o: o.a + 1),
          lambda o: o.b)
  gives: 2
  '''
  if len(bindings) == 0:
    return body(env)

  env = env or _obj()
  k, v = bindings[:2]
  if isinstance(v, types.FunctionType):
    v = v(env)

  setattr(env, k, v)
  return LET(bindings[2:], body, env)

You can now use this LET form as such:

map(lambda x: LET(('_', x.sort()),
                  lambda _: x[1]),
    lst)

which gives: [345, 465, 333]


回答 9

您可以使用min和index而不是使用sort或heapq在O(n)时间内完成此操作。

首先创建除原始列表的最小值以外的所有内容的新列表:

new_list = lst[:lst.index(min(lst))] + lst[lst.index(min(lst))+1:]

然后取新列表的最小值:

second_smallest = min(new_list)

现在,所有这些都放在一个lambda中:

map(lambda x: min(x[:x.index(min(x))] + x[x.index(min(x))+1:]), lst)

是的,它确实很丑陋,但是在算法上应该便宜。同样由于该线程中的某些人希望查看列表理解:

[min(x[:x.index(min(x))] + x[x.index(min(x))+1:]) for x in lst]

You can do it in O(n) time using min and index instead of using sort or heapq.

First create new list of everything except the min value of the original list:

new_list = lst[:lst.index(min(lst))] + lst[lst.index(min(lst))+1:]

Then take the min value of the new list:

second_smallest = min(new_list)

Now all together in a single lambda:

map(lambda x: min(x[:x.index(min(x))] + x[x.index(min(x))+1:]), lst)

Yes it is really ugly, but it should be algorithmically cheap. Also since some folks in this thread want to see list comprehensions:

[min(x[:x.index(min(x))] + x[x.index(min(x))+1:]) for x in lst]

回答 10

这也正是bind在函数单子的用途。

使用此bind功能,您可以将多个lambda组合为一个lambda,每个lambda代表一条语句。

This is exactly what the bind function in a Monad is used for.

With the bind function you can combine multiple lambda’s into one lambda, each lambda representing a statement.


回答 11

实际上,有一种方法可以在lambda中使用多个语句。这是我的解决方案:

lst = [[567,345,234],[253,465,756, 2345],[333,777,111, 555]]

x = lambda l: exec("l.sort(); return l[1]")

map(x, lst)

There actually is a way you can use multiple statements in lambda. Here’s my solution:

lst = [[567,345,234],[253,465,756, 2345],[333,777,111, 555]]

x = lambda l: exec("l.sort(); return l[1]")

map(x, lst)

回答 12

是。您可以用这种方式定义它,然后用以下代码包装多个表达式:

计划开始:

开始= lambda * x:x [-1]

Lisp常见问题:

progn = lambda * x:x [-1]

Yes. You can define it this way and then wrap your multiple expressions with the following:

Scheme begin:

begin = lambda *x: x[-1]

Common Lisp progn:

progn = lambda *x: x[-1]


回答 13

没有使用lambda函数的更好的解决方案。但是,如果我们真的想使用lambda函数,则这是处理多个语句的通用解决方案:map(lambda x:x [1] if(x.sort())else x [1],lst)

您实际上并不关心语句返回的内容。

There are better solutions without using lambda function. But if we really want to use lambda function, here is a generic solution to deal with multiple statements: map(lambda x: x[1] if (x.sort()) else x[1],lst)

You don’t really care what the statement returns.


回答 14

是的,有可能。请尝试以下代码段。

x = [('human', 1), ('i', 2), ('am', 1), ('.', 1), ('love', 1), ('python', 3), ('', 1),
  ('run', 1), ('is', 2), ('robust', 1), ('hello', 1), ('spark', 2), ('to', 1), ('analysis', 2), ('on', 1), ('big', 1), ('data', 1), ('with', 1), ('analysis', 1), ('great', 1)
]

rdd_filter = rdd1_word_cnt_sum.filter(lambda x: 'python' in x or 'human' in x or 'big' in x)
rdd_filter.collect()

Yes it is possible. Try below code snippet.

x = [('human', 1), ('i', 2), ('am', 1), ('.', 1), ('love', 1), ('python', 3), ('', 1),
  ('run', 1), ('is', 2), ('robust', 1), ('hello', 1), ('spark', 2), ('to', 1), ('analysis', 2), ('on', 1), ('big', 1), ('data', 1), ('with', 1), ('analysis', 1), ('great', 1)
]

rdd_filter = rdd1_word_cnt_sum.filter(lambda x: 'python' in x or 'human' in x or 'big' in x)
rdd_filter.collect()

回答 15

我将给您另一个解决方案,使您的lambda调用一个函数。

def multiple_statements(x, y):
    print('hi')
    print('there')
    print(x)
    print(y)
    return 1

junky = lambda x, y: multiple_statements(x, y)

junky('a', 'b');

I’ll give you another solution, Make your lambda invoke a function.

def multiple_statements(x, y):
    print('hi')
    print('there')
    print(x)
    print(y)
    return 1

junky = lambda x, y: multiple_statements(x, y)

junky('a', 'b');