标签归档:reduce

函数式编程中的“ pythonic”等同于“ fold”函数是什么?

问题:函数式编程中的“ pythonic”等同于“ fold”函数是什么?

在Haskell中,实现以下目标的最惯用的方法是:

foldl (+) 0 [1,2,3,4,5]
--> 15

或等效的Ruby:

[1,2,3,4,5].inject(0) {|m,x| m + x}
#> 15

显然,Python提供了reduce与fold完全相同的功能,但实际上是如上所述的fold的实现,但是,有人告诉我,“ pythonic”编程方式是避免使用lambda术语和高阶函数,并尽可能使用列表理解。因此,有没有一种首选的方式来折叠列表或不是Python reduce函数的类似列表的结构,或者是reduce惯用的方式来实现此目的?

What is the most idiomatic way to achieve something like the following, in Haskell:

foldl (+) 0 [1,2,3,4,5]
--> 15

Or its equivalent in Ruby:

[1,2,3,4,5].inject(0) {|m,x| m + x}
#> 15

Obviously, Python provides the reduce function, which is an implementation of fold, exactly as above, however, I was told that the ‘pythonic’ way of programming was to avoid lambda terms and higher-order functions, preferring list-comprehensions where possible. Therefore, is there a preferred way of folding a list, or list-like structure in Python that isn’t the reduce function, or is reduce the idiomatic way of achieving this?


回答 0

用Python方式对数组求和的方法是使用sum。为了其他目的,有时可以使用reduce(来自functools模块)和operator模块的某种组合,例如:

def product(xs):
    return reduce(operator.mul, xs, 1)

请注意,reduce实际上这是一个foldl用Haskell表示的。没有执行折叠的特殊语法,没有内置函数foldr,实际上reduce与非关联运算符一起使用被认为是不良样式。

使用高阶函数是相当Python的;它很好地利用了Python的原理,即一切都是对象,包括函数和类。没错,lambda被某些Pythonista所反对,但这主要是因为它们变得复杂时往往不太可读。

The Pythonic way of summing an array is using sum. For other purposes, you can sometimes use some combination of reduce (from the functools module) and the operator module, e.g.:

def product(xs):
    return reduce(operator.mul, xs, 1)

Be aware that reduce is actually a foldl, in Haskell terms. There is no special syntax to perform folds, there’s no builtin foldr, and actually using reduce with non-associative operators is considered bad style.

Using higher-order functions is quite pythonic; it makes good use of Python’s principle that everything is an object, including functions and classes. You are right that lambdas are frowned upon by some Pythonistas, but mostly because they tend not to be very readable when they get complex.


回答 1

哈斯克尔

foldl (+) 0 [1,2,3,4,5]

Python

reduce(lambda a,b: a+b, [1,2,3,4,5], 0)

显然,这是一个说明问题的简单例子。在Python中,您只需要这样做sum([1,2,3,4,5]),甚至Haskell纯粹主义者通常也会更喜欢sum [1,2,3,4,5]

对于没有明显便利函数的非平凡场景,惯用的pythonic方法是显式写出for循环并使用可变变量分配,而不是使用reduceor fold

那根本不是功能样式,而是“ pythonic”方式。Python并非为功能纯正者设计。了解Python如何支持流控制的异常,以了解非功能性惯用python的情况。

Haskell

foldl (+) 0 [1,2,3,4,5]

Python

reduce(lambda a,b: a+b, [1,2,3,4,5], 0)

Obviously, that is a trivial example to illustrate a point. In Python you would just do sum([1,2,3,4,5]) and even Haskell purists would generally prefer sum [1,2,3,4,5].

For non-trivial scenarios when there is no obvious convenience function, the idiomatic pythonic approach is to explicitly write out the for loop and use mutable variable assignment instead of using reduce or a fold.

That is not at all the functional style, but that is the “pythonic” way. Python is not designed for functional purists. See how Python favors exceptions for flow control to see how non-functional idiomatic python is.


回答 2

在Python 3中,reduce已被删除:版本说明。不过,您可以使用functools模块

import operator, functools
def product(xs):
    return functools.reduce(operator.mul, xs, 1)

另一方面,文档表达了对for-loop而不是的偏好reduce,因此:

def product(xs):
    result = 1
    for i in xs:
        result *= i
    return result

In Python 3, the reduce has been removed: Release notes. Nevertheless you can use the functools module

import operator, functools
def product(xs):
    return functools.reduce(operator.mul, xs, 1)

On the other hand, the documentation expresses preference towards for-loop instead of reduce, hence:

def product(xs):
    result = 1
    for i in xs:
        result *= i
    return result

回答 3

您也可以重新发明轮子:

def fold(f, l, a):
    """
    f: the function to apply
    l: the list to fold
    a: the accumulator, who is also the 'zero' on the first call
    """ 
    return a if(len(l) == 0) else fold(f, l[1:], f(a, l[0]))

print "Sum:", fold(lambda x, y : x+y, [1,2,3,4,5], 0)

print "Any:", fold(lambda x, y : x or y, [False, True, False], False)

print "All:", fold(lambda x, y : x and y, [False, True, False], True)

# Prove that result can be of a different type of the list's elements
print "Count(x==True):", 
print fold(lambda x, y : x+1 if(y) else x, [False, True, True], 0)

You can reinvent the wheel as well:

def fold(f, l, a):
    """
    f: the function to apply
    l: the list to fold
    a: the accumulator, who is also the 'zero' on the first call
    """ 
    return a if(len(l) == 0) else fold(f, l[1:], f(a, l[0]))

print "Sum:", fold(lambda x, y : x+y, [1,2,3,4,5], 0)

print "Any:", fold(lambda x, y : x or y, [False, True, False], False)

print "All:", fold(lambda x, y : x and y, [False, True, False], True)

# Prove that result can be of a different type of the list's elements
print "Count(x==True):", 
print fold(lambda x, y : x+1 if(y) else x, [False, True, True], 0)

回答 4

并不是真正回答这个问题,而是折叠和折叠的一线:

a = [8,3,4]

## Foldl
reduce(lambda x,y: x**y, a)
#68719476736

## Foldr
reduce(lambda x,y: y**x, a[::-1])
#14134776518227074636666380005943348126619871175004951664972849610340958208L

Not really answer to the question, but one-liners for foldl and foldr:

a = [8,3,4]

## Foldl
reduce(lambda x,y: x**y, a)
#68719476736

## Foldr
reduce(lambda x,y: y**x, a[::-1])
#14134776518227074636666380005943348126619871175004951664972849610340958208L

回答 5

开始Python 3.8并引入赋值表达式(PEP 572):=运算符),这使您可以为表达式的结果命名,我们可以使用列表推导来复制其他语言称为fold / foldleft / reduce的操作:

给定一个列表,一个约简函数和一个累加器:

items = [1, 2, 3, 4, 5]
f = lambda acc, x: acc * x
accumulator = 1

我们可以折叠itemsf在为了获得所得accumulation

[accumulator := f(accumulator, x) for x in items]
# accumulator = 120

或以压缩形式形成:

acc = 1; [acc := acc * x for x in [1, 2, 3, 4, 5]]
# acc = 120

请注意,这实际上也是“ scanleft”操作,因为列表理解的结果表示每个步骤的累加状态:

acc = 1
scanned = [acc := acc * x for x in [1, 2, 3, 4, 5]]
# scanned = [1, 2, 6, 24, 120]
# acc = 120

Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), which gives the possibility to name the result of an expression, we can use a list comprehension to replicate what other languages call fold/foldleft/reduce operations:

Given a list, a reducing function and an accumulator:

items = [1, 2, 3, 4, 5]
f = lambda acc, x: acc * x
accumulator = 1

we can fold items with f in order to obtain the resulting accumulation:

[accumulator := f(accumulator, x) for x in items]
# accumulator = 120

or in a condensed formed:

acc = 1; [acc := acc * x for x in [1, 2, 3, 4, 5]]
# acc = 120

Note that this is actually also a “scanleft” operation as the result of the list comprehension represents the state of the accumulation at each step:

acc = 1
scanned = [acc := acc * x for x in [1, 2, 3, 4, 5]]
# scanned = [1, 2, 6, 24, 120]
# acc = 120

回答 6

对这个(减少)问题的实际答案是:只需使用一个循环!

initial_value = 0
for x in the_list:
    initial_value += x #or any function.

这将比reduce更快,而PyPy之类的东西可以优化循环。

顺便说一句,总和的情况应该用sum函数来解决

The actual answer to this (reduce) problem is: Just use a loop!

initial_value = 0
for x in the_list:
    initial_value += x #or any function.

This will be faster than a reduce and things like PyPy can optimize loops like that.

BTW, the sum case should be solved with the sum function


回答 7

我相信这个问题的一些回答者错过了该fold功能作为抽象工具的广泛含义。是的,sum可以对整数列表执行相同的操作,但这是不重要的情况。fold更通用。当您具有一系列形状各异的数据结构并想要清晰地表达聚合时,此功能很有用。因此,不必for每次都用一个聚合变量建立一个循环并手动重新计算它,fold函数(或reduce似乎对应的Python版本)允许程序员通过简单地提供以下内容来更清楚地表达聚合的意图:两件事情:

  • 聚合的默认起始值​​或“种子”值。
  • 该函数采用聚合的当前值(以“种子”开头)和列表中的下一个元素,并返回下一个聚合值。

I believe some of the respondents of this question have missed the broader implication of the fold function as an abstract tool. Yes, sum can do the same thing for a list of integers, but this is a trivial case. fold is more generic. It is useful when you have a sequence of data structures of varying shape and want to cleanly express an aggregation. So instead of having to build up a for loop with an aggregate variable and manually recompute it each time, a fold function (or the Python version, which reduce appears to correspond to) allows the programmer to express the intent of the aggregation much more plainly by simply providing two things:

  • A default starting or “seed” value for the aggregation.
  • A function that takes the current value of the aggregation (starting with the “seed”) and the next element in the list, and returns the next aggregation value.

回答 8

我参加聚会可能已经很晚了,但是我们可以foldr使用简单的lambda演算和curried函数创建自定义项。这是我在python中实现的foldr。

def foldr(func):
    def accumulator(acc):
        def listFunc(l):
            if l:
                x = l[0]
                xs = l[1:]
                return func(x)(foldr(func)(acc)(xs))
            else:
                return acc
        return listFunc
    return accumulator  


def curried_add(x):
    def inner(y):
        return x + y
    return inner

def curried_mult(x):
    def inner(y):
        return x * y
    return inner

print foldr(curried_add)(0)(range(1, 6))
print foldr(curried_mult)(1)(range(1, 6))

即使实现是递归的(可能很慢),它也会分别打印值15120

I may be quite late to the party, but we can create custom foldr using simple lambda calculus and curried function. Here is my implementation of foldr in python.

def foldr(func):
    def accumulator(acc):
        def listFunc(l):
            if l:
                x = l[0]
                xs = l[1:]
                return func(x)(foldr(func)(acc)(xs))
            else:
                return acc
        return listFunc
    return accumulator  


def curried_add(x):
    def inner(y):
        return x + y
    return inner

def curried_mult(x):
    def inner(y):
        return x * y
    return inner

print foldr(curried_add)(0)(range(1, 6))
print foldr(curried_mult)(1)(range(1, 6))

Even though the implementation is recursive (might be slow), it will print the values 15 and 120 respectively


NameError:名称“ reduce”未在Python中定义

问题:NameError:名称“ reduce”未在Python中定义

我正在使用Python 3.2。试过这个:

xor = lambda x,y: (x+y)%2
l = reduce(xor, [1,2,3,4])

并得到以下错误:

l = reduce(xor, [1,2,3,4])
NameError: name 'reduce' is not defined

尝试打印reduce到交互式控制台中-出现此错误:

NameError: name 'reduce' is not defined


reduce在Python 3.2中真正删除的吗?如果是这样,还有什么选择?

I’m using Python 3.2. Tried this:

xor = lambda x,y: (x+y)%2
l = reduce(xor, [1,2,3,4])

And got the following error:

l = reduce(xor, [1,2,3,4])
NameError: name 'reduce' is not defined

Tried printing reduce into interactive console – got this error:

NameError: name 'reduce' is not defined


Is reduce really removed in Python 3.2? If that’s the case, what’s the alternative?


回答 0

它已移至functools


回答 1

你可以加

from functools import reduce

在使用reduce之前。

You can add

from functools import reduce

before you use the reduce.


回答 2

或者如果您使用六个库

from six.moves import reduce

Or if you use the six library

from six.moves import reduce

回答 3

在这种情况下,我认为以下内容是等效的:

l = sum([1,2,3,4]) % 2

唯一的问题是它会产生大量数字,但这也许比重复的模运算更好?

In this case I believe that the following is equivalent:

l = sum([1,2,3,4]) % 2

The only problem with this is that it creates big numbers, but maybe that is better than repeated modulo operations?


如何在Python 3中使用过滤,映射和归约

问题:如何在Python 3中使用过滤,映射和归约

filter,,map并且reduce可以在Python 2中完美运行。这是一个示例:

>>> def f(x):
        return x % 2 != 0 and x % 3 != 0
>>> filter(f, range(2, 25))
[5, 7, 11, 13, 17, 19, 23]

>>> def cube(x):
        return x*x*x
>>> map(cube, range(1, 11))
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

>>> def add(x,y):
        return x+y
>>> reduce(add, range(1, 11))
55

但是在Python 3中,我收到以下输出:

>>> filter(f, range(2, 25))
<filter object at 0x0000000002C14908>

>>> map(cube, range(1, 11))
<map object at 0x0000000002C82B70>

>>> reduce(add, range(1, 11))
Traceback (most recent call last):
  File "<pyshell#8>", line 1, in <module>
    reduce(add, range(1, 11))
NameError: name 'reduce' is not defined

如果有人可以向我解释为什么,我将不胜感激。

代码的屏幕截图,用于进一步说明:

filter, map, and reduce work perfectly in Python 2. Here is an example:

>>> def f(x):
        return x % 2 != 0 and x % 3 != 0
>>> filter(f, range(2, 25))
[5, 7, 11, 13, 17, 19, 23]

>>> def cube(x):
        return x*x*x
>>> map(cube, range(1, 11))
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

>>> def add(x,y):
        return x+y
>>> reduce(add, range(1, 11))
55

But in Python 3, I receive the following outputs:

>>> filter(f, range(2, 25))
<filter object at 0x0000000002C14908>

>>> map(cube, range(1, 11))
<map object at 0x0000000002C82B70>

>>> reduce(add, range(1, 11))
Traceback (most recent call last):
  File "<pyshell#8>", line 1, in <module>
    reduce(add, range(1, 11))
NameError: name 'reduce' is not defined

I would appreciate if someone could explain to me why this is.

Screenshot of code for further clarity:


回答 0

您可以阅读Python 3.0的新增功能中的更改。从2.x升级到3.x时,应该仔细阅读它,因为已经做了很多更改。

此处的完整答案是文档中的引号。

视图和迭代器而不是列表

一些著名的API不再返回列表:

  • […]
  • map()filter()返回迭代器。如果您确实需要列表,则可以使用快速解决方案,例如list(map(...)),但是更好的解决方案通常是使用列表理解(特别是当原始代码使用lambda时),或者重写代码以使其根本不需要列表。map()该函数的副作用特别棘手。正确的转换是使用常规for循环(因为创建列表将很浪费)。
  • […]

内建

  • […]
  • 已删除reduce()functools.reduce()如果确实需要,请使用;但是,在99%的时间里,显式for循环更易于阅读。
  • […]

You can read about the changes in What’s New In Python 3.0. You should read it thoroughly when you move from 2.x to 3.x since a lot has been changed.

The whole answer here are quotes from the documentation.

Views And Iterators Instead Of Lists

Some well-known APIs no longer return lists:

  • […]
  • map() and filter() return iterators. If you really need a list, a quick fix is e.g. list(map(...)), but a better fix is often to use a list comprehension (especially when the original code uses lambda), or rewriting the code so it doesn’t need a list at all. Particularly tricky is map() invoked for the side effects of the function; the correct transformation is to use a regular for loop (since creating a list would just be wasteful).
  • […]

Builtins

  • […]
  • Removed reduce(). Use functools.reduce() if you really need it; however, 99 percent of the time an explicit for loop is more readable.
  • […]

回答 1

的功能mapfilter被有意改为返回迭代器,并减少从被除去的内置和放置在functools.reduce

因此,对于filtermap,您可以将它们包装起来以list()像以前一样查看结果。

>>> def f(x): return x % 2 != 0 and x % 3 != 0
...
>>> list(filter(f, range(2, 25)))
[5, 7, 11, 13, 17, 19, 23]
>>> def cube(x): return x*x*x
...
>>> list(map(cube, range(1, 11)))
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
>>> import functools
>>> def add(x,y): return x+y
...
>>> functools.reduce(add, range(1, 11))
55
>>>

现在的建议是,用生成器表达式或列表推导替换map和filter的用法。例:

>>> def f(x): return x % 2 != 0 and x % 3 != 0
...
>>> [i for i in range(2, 25) if f(i)]
[5, 7, 11, 13, 17, 19, 23]
>>> def cube(x): return x*x*x
...
>>> [cube(i) for i in range(1, 11)]
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
>>>

他们说for循环在99%的时间里比减少容易阅读,但是我还是坚持functools.reduce

编辑:99%的数字直接从Guido van Rossum编写的“ Python 3.0的新功能”页面中提取。

The functionality of map and filter was intentionally changed to return iterators, and reduce was removed from being a built-in and placed in functools.reduce.

So, for filter and map, you can wrap them with list() to see the results like you did before.

>>> def f(x): return x % 2 != 0 and x % 3 != 0
...
>>> list(filter(f, range(2, 25)))
[5, 7, 11, 13, 17, 19, 23]
>>> def cube(x): return x*x*x
...
>>> list(map(cube, range(1, 11)))
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
>>> import functools
>>> def add(x,y): return x+y
...
>>> functools.reduce(add, range(1, 11))
55
>>>

The recommendation now is that you replace your usage of map and filter with generators expressions or list comprehensions. Example:

>>> def f(x): return x % 2 != 0 and x % 3 != 0
...
>>> [i for i in range(2, 25) if f(i)]
[5, 7, 11, 13, 17, 19, 23]
>>> def cube(x): return x*x*x
...
>>> [cube(i) for i in range(1, 11)]
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
>>>

They say that for loops are 99 percent of the time easier to read than reduce, but I’d just stick with functools.reduce.

Edit: The 99 percent figure is pulled directly from the What’s New In Python 3.0 page authored by Guido van Rossum.


回答 2

作为其他答案的补充,对于上下文管理器来说,这听起来像是一个很好的用例,它将重新将这些函数的名称映射为返回列表并引入reduce全局命名空间的函数。

快速实现可能如下所示:

from contextlib import contextmanager    

@contextmanager
def noiters(*funcs):
    if not funcs: 
        funcs = [map, filter, zip] # etc
    from functools import reduce
    globals()[reduce.__name__] = reduce
    for func in funcs:
        globals()[func.__name__] = lambda *ar, func = func, **kwar: list(func(*ar, **kwar))
    try:
        yield
    finally:
        del globals()[reduce.__name__]
        for func in funcs: globals()[func.__name__] = func

用法如下所示:

with noiters(map):
    from operator import add
    print(reduce(add, range(1, 20)))
    print(map(int, ['1', '2']))

哪些打印:

190
[1, 2]

只是我的2美分:-)

As an addendum to the other answers, this sounds like a fine use-case for a context manager that will re-map the names of these functions to ones which return a list and introduce reduce in the global namespace.

A quick implementation might look like this:

from contextlib import contextmanager    

@contextmanager
def noiters(*funcs):
    if not funcs: 
        funcs = [map, filter, zip] # etc
    from functools import reduce
    globals()[reduce.__name__] = reduce
    for func in funcs:
        globals()[func.__name__] = lambda *ar, func = func, **kwar: list(func(*ar, **kwar))
    try:
        yield
    finally:
        del globals()[reduce.__name__]
        for func in funcs: globals()[func.__name__] = func

With a usage that looks like this:

with noiters(map):
    from operator import add
    print(reduce(add, range(1, 20)))
    print(map(int, ['1', '2']))

Which prints:

190
[1, 2]

Just my 2 cents 🙂


回答 3

由于该reduce方法已从Python3的内置函数中删除,因此请不要忘记functools在您的代码中导入。请查看下面的代码段。

import functools
my_list = [10,15,20,25,35]
sum_numbers = functools.reduce(lambda x ,y : x+y , my_list)
print(sum_numbers)

Since the reduce method has been removed from the built in function from Python3, don’t forget to import the functools in your code. Please look at the code snippet below.

import functools
my_list = [10,15,20,25,35]
sum_numbers = functools.reduce(lambda x ,y : x+y , my_list)
print(sum_numbers)

回答 4

以下是Filter,map和reduce函数的示例。

数字= [10,11,12,22,34,43,54,34,67,87,88,98,99,87,44,66]

//过滤

奇数=列表(filter(lambda x:x%2!= 0,数字))

打印(奇数)

//地图

multipleOf2 = list(map(lambda x:x * 2,数字))

打印(multiplyOf2)

//降低

由于不常用reduce函数,因此已从Python 3的内置函数中删除了它。functools模块中仍提供了reduce函数,因此您可以执行以下操作:

从functools进口减少

sumOfNumbers = reduce(lambda x,y:x + y,数字)

打印(sumOfNumbers)

Here are the examples of Filter, map and reduce functions.

numbers = [10,11,12,22,34,43,54,34,67,87,88,98,99,87,44,66]

//Filter

oddNumbers = list(filter(lambda x: x%2 != 0, numbers))

print(oddNumbers)

//Map

multiplyOf2 = list(map(lambda x: x*2, numbers))

print(multiplyOf2)

//Reduce

The reduce function, since it is not commonly used, was removed from the built-in functions in Python 3. It is still available in the functools module, so you can do:

from functools import reduce

sumOfNumbers = reduce(lambda x,y: x+y, numbers)

print(sumOfNumbers)


回答 5

map,filter和reduce的优点之一是当您将它们“链接”在一起以进行复杂的操作时,它们变得清晰易读。但是,内置语法不清晰,全都是“向后的”。因此,我建议使用该PyFunctional软件包(https://pypi.org/project/PyFunctional/)。 这是两者的比较:

flight_destinations_dict = {'NY': {'London', 'Rome'}, 'Berlin': {'NY'}}

Py功能版本

非常清晰的语法。你可以说:

“我有一个飞行目的地序列。如果城市位于dict值中,我想从中获得dict键。最后,过滤掉我在流程中创建的空列表。”

from functional import seq  # PyFunctional package to allow easier syntax

def find_return_flights_PYFUNCTIONAL_SYNTAX(city, flight_destinations_dict):
    return seq(flight_destinations_dict.items()) \
        .map(lambda x: x[0] if city in x[1] else []) \
        .filter(lambda x: x != []) \

默认Python版本

都是倒退。您需要说:

“好的,所以有一个列表。我想从中过滤出空列表。为什么?因为如果城市位于dict值中,我首先得到了dict键。哦,我要执行的列表是flight_destinations_dict。 ”

def find_return_flights_DEFAULT_SYNTAX(city, flight_destinations_dict):
    return list(
        filter(lambda x: x != [],
               map(lambda x: x[0] if city in x[1] else [], flight_destinations_dict.items())
               )
    )

One of the advantages of map, filter and reduce is how legible they become when you “chain” them together to do something complex. However, the built-in syntax isn’t legible and is all “backwards”. So, I suggest using the PyFunctional package (https://pypi.org/project/PyFunctional/). Here’s a comparison of the two:

flight_destinations_dict = {'NY': {'London', 'Rome'}, 'Berlin': {'NY'}}

PyFunctional version

Very legible syntax. You can say:

“I have a sequence of flight destinations. Out of which I want to get the dict key if city is in the dict values. Finally, filter out the empty lists I created in the process.”

from functional import seq  # PyFunctional package to allow easier syntax

def find_return_flights_PYFUNCTIONAL_SYNTAX(city, flight_destinations_dict):
    return seq(flight_destinations_dict.items()) \
        .map(lambda x: x[0] if city in x[1] else []) \
        .filter(lambda x: x != []) \

Default Python version

It’s all backwards. You need to say:

“OK, so, there’s a list. I want to filter empty lists out of it. Why? Because I first got the dict key if the city was in the dict values. Oh, the list I’m doing this to is flight_destinations_dict.”

def find_return_flights_DEFAULT_SYNTAX(city, flight_destinations_dict):
    return list(
        filter(lambda x: x != [],
               map(lambda x: x[0] if city in x[1] else [], flight_destinations_dict.items())
               )
    )

查找列表的平均值

问题:查找列表的平均值

我必须在Python中找到列表的平均值。到目前为止,这是我的代码

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print reduce(lambda x, y: x + y, l)

我已经知道了,所以它可以将列表中的值加在一起,但是我不知道如何将其划分为它们?

I have to find the average of a list in Python. This is my code so far

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print reduce(lambda x, y: x + y, l)

I’ve got it so it adds together the values in the list, but I don’t know how to make it divide into them?


回答 0

在Python 3.4+上,您可以使用 statistics.mean()

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import statistics
statistics.mean(l)  # 20.11111111111111

在旧版本的Python上,您可以执行

sum(l) / len(l)

在Python 2上,您需要转换len为浮点数才能进行浮点数除法

sum(l) / float(len(l))

无需使用reduce。它慢得多,并在Python 3 中删除

On Python 3.4+ you can use statistics.mean()

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import statistics
statistics.mean(l)  # 20.11111111111111

On older versions of Python you can do

sum(l) / len(l)

On Python 2 you need to convert len to a float to get float division

sum(l) / float(len(l))

There is no need to use reduce. It is much slower and was removed in Python 3.


回答 1

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(l) / len(l)
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(l) / len(l)

回答 2

您可以使用numpy.mean

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import numpy as np
print(np.mean(l))

You can use numpy.mean:

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import numpy as np
print(np.mean(l))

回答 3

一个统计模块已经加入到了Python 3.4。它具有计算平均值的功能,称为均值。您提供的列表的示例为:

from statistics import mean
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
mean(l)

A statistics module has been added to python 3.4. It has a function to calculate the average called mean. An example with the list you provided would be:

from statistics import mean
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
mean(l)

回答 4

reduce()当Python具有完美的sum()功能时,为什么要使用此功能?

print sum(l) / float(len(l))

float()必须强制Python执行浮点除法。)

Why would you use reduce() for this when Python has a perfectly cromulent sum() function?

print sum(l) / float(len(l))

(The float() is necessary to force Python to do a floating-point division.)


回答 5

如果您使用的是python> = 3.4,则有一个统计资料库

https://docs.python.org/3/library/statistics.html

您可以使用这种卑鄙的方法。假设您有一个要查找均值的数字列表:-

list = [11, 13, 12, 15, 17]
import statistics as s
s.mean(list)

它还有其他方法,如stdev,方差,众数,谐波均值,中位数等,这些方法也非常有用。

There is a statistics library if you are using python >= 3.4

https://docs.python.org/3/library/statistics.html

You may use it’s mean method like this. Let’s say you have a list of numbers of which you want to find mean:-

list = [11, 13, 12, 15, 17]
import statistics as s
s.mean(list)

It has other methods too like stdev, variance, mode, harmonic mean, median etc which are too useful.


回答 6

除了将其强制转换为浮点数外,还可以在总和上加上0.0:

def avg(l):
    return sum(l, 0.0) / len(l)

Instead of casting to float, you can add 0.0 to the sum:

def avg(l):
    return sum(l, 0.0) / len(l)

回答 7

sum(l) / float(len(l)) 是正确的答案,但仅出于完整性考虑,您可以通过一次减少来计算平均值:

>>> reduce(lambda x, y: x + y / float(len(l)), l, 0)
20.111111111111114

请注意,这可能会导致轻微的舍入错误:

>>> sum(l) / float(len(l))
20.111111111111111

sum(l) / float(len(l)) is the right answer, but just for completeness you can compute an average with a single reduce:

>>> reduce(lambda x, y: x + y / float(len(l)), l, 0)
20.111111111111114

Note that this can result in a slight rounding error:

>>> sum(l) / float(len(l))
20.111111111111111

回答 8

我尝试使用上面的选项,但是没有用。尝试这个:

from statistics import mean

n = [11, 13, 15, 17, 19]

print(n)
print(mean(n))

在python 3.5上工作

I tried using the options above but didn’t work. Try this:

from statistics import mean

n = [11, 13, 15, 17, 19]

print(n)
print(mean(n))

worked on python 3.5


回答 9

或使用pandasSeries.mean方法:

pd.Series(sequence).mean()

演示:

>>> import pandas as pd
>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> pd.Series(l).mean()
20.11111111111111
>>> 

从文档:

Series.mean(axis=None, skipna=None, level=None, numeric_only=None, **kwargs)

这是文档:

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.mean.html

以及整个文档:

https://pandas.pydata.org/pandas-docs/stable/10min.html

Or use pandas‘s Series.mean method:

pd.Series(sequence).mean()

Demo:

>>> import pandas as pd
>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> pd.Series(l).mean()
20.11111111111111
>>> 

From the docs:

Series.mean(axis=None, skipna=None, level=None, numeric_only=None, **kwargs)

And here is the docs for this:

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.mean.html

And the whole documentation:

https://pandas.pydata.org/pandas-docs/stable/10min.html


回答 10

在Udacity的问题中,我也有类似的问题要解决。我编码的不是内置函数:

def list_mean(n):

    summing = float(sum(n))
    count = float(len(n))
    if n == []:
        return False
    return float(summing/count)

比平常更长的时间,但是对于初学者来说,这是一个很大的挑战。

I had a similar question to solve in a Udacity´s problems. Instead of a built-in function i coded:

def list_mean(n):

    summing = float(sum(n))
    count = float(len(n))
    if n == []:
        return False
    return float(summing/count)

Much more longer than usual but for a beginner its quite challenging.


回答 11

作为一个初学者,我只是编写了这样的代码:

L = [15, 18, 2, 36, 12, 78, 5, 6, 9]

total = 0

def average(numbers):
    total = sum(numbers)
    total = float(total)
    return total / len(numbers)

print average(L)

as a beginner, I just coded this:

L = [15, 18, 2, 36, 12, 78, 5, 6, 9]

total = 0

def average(numbers):
    total = sum(numbers)
    total = float(total)
    return total / len(numbers)

print average(L)

回答 12

如果您想获得的不仅仅是平均值(也就是平均值),您可以查看scipy统计信息

from scipy import stats
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(stats.describe(l))

# DescribeResult(nobs=9, minmax=(2, 78), mean=20.11111111111111, 
# variance=572.3611111111111, skewness=1.7791785448425341, 
# kurtosis=1.9422716419666397)

If you wanted to get more than just the mean (aka average) you might check out scipy stats

from scipy import stats
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(stats.describe(l))

# DescribeResult(nobs=9, minmax=(2, 78), mean=20.11111111111111, 
# variance=572.3611111111111, skewness=1.7791785448425341, 
# kurtosis=1.9422716419666397)

回答 13

为了reduce用于获得运行平均值,您需要跟踪总数,但也要跟踪到目前为止看到的元素总数。由于这不是列表中的琐碎元素,因此您还必须传递reduce一个额外的参数以使其折叠。

>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> running_average = reduce(lambda aggr, elem: (aggr[0] + elem, aggr[1]+1), l, (0.0,0))
>>> running_average[0]
(181.0, 9)
>>> running_average[0]/running_average[1]
20.111111111111111

In order to use reduce for taking a running average, you’ll need to track the total but also the total number of elements seen so far. since that’s not a trivial element in the list, you’ll also have to pass reduce an extra argument to fold into.

>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> running_average = reduce(lambda aggr, elem: (aggr[0] + elem, aggr[1]+1), l, (0.0,0))
>>> running_average[0]
(181.0, 9)
>>> running_average[0]/running_average[1]
20.111111111111111

回答 14

两者都可以为您提供接近整数或至少10个十进制值的相似值。但是,如果您真正考虑的是长浮点值,则两者可能会有所不同。方法可能因您要实现的目标而异。

>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> print reduce(lambda x, y: x + y, l) / len(l)
20
>>> sum(l)/len(l)
20

浮动值

>>> print reduce(lambda x, y: x + y, l) / float(len(l))
20.1111111111
>>> print sum(l)/float(len(l))
20.1111111111

@安德鲁·克拉克(Andrew Clark)的发言是正确的。

Both can give you close to similar values on an integer or at least 10 decimal values. But if you are really considering long floating values both can be different. Approach can vary on what you want to achieve.

>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> print reduce(lambda x, y: x + y, l) / len(l)
20
>>> sum(l)/len(l)
20

Floating values

>>> print reduce(lambda x, y: x + y, l) / float(len(l))
20.1111111111
>>> print sum(l)/float(len(l))
20.1111111111

@Andrew Clark was correct on his statement.


回答 15

假设

x = [[-5.01,-5.43,1.08,0.86,-2.67,4.94,-2.51,-2.25,5.56,1.03], [-8.12,-3.48,-5.52,-3.78,0.63,3.29,2.09,-2.13,2.86,-3.33], [-3.68,-3.54,1.66,-4.11,7.39,2.08,-2.59,-6.94,-2.26,4.33]]

您会注意到它的x尺寸为3 * 10,如果您需要mean进入每一行,则可以键入

theMean = np.mean(x1,axis=1)

别忘了 import numpy as np

suppose that

x = [[-5.01,-5.43,1.08,0.86,-2.67,4.94,-2.51,-2.25,5.56,1.03], [-8.12,-3.48,-5.52,-3.78,0.63,3.29,2.09,-2.13,2.86,-3.33], [-3.68,-3.54,1.66,-4.11,7.39,2.08,-2.59,-6.94,-2.26,4.33]]

you can notice that x has dimension 3*10 if you need to get the mean to each row you can type this

theMean = np.mean(x1,axis=1)

don’t forget to import numpy as np


回答 16

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

l = map(float,l)
print '%.2f' %(sum(l)/len(l))
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

l = map(float,l)
print '%.2f' %(sum(l)/len(l))

回答 17

使用以下PYTHON代码在列表中查找平均值:

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(sum(l)//len(l))

试试这个很容易。

Find the average in list By using the following PYTHON code:

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(sum(l)//len(l))

try this it easy.


回答 18

print reduce(lambda x, y: x + y, l)/(len(l)*1.0)

或喜欢以前发布的

sum(l)/(len(l)*1.0)

1.0是确保获得浮点除法

print reduce(lambda x, y: x + y, l)/(len(l)*1.0)

or like posted previously

sum(l)/(len(l)*1.0)

The 1.0 is to make sure you get a floating point division


回答 19

结合以上几个答案,我提出了以下与reduce一起使用的方法,并且不假定您在reduce L函数中可用:

from operator import truediv

L = [15, 18, 2, 36, 12, 78, 5, 6, 9]

def sum_and_count(x, y):
    try:
        return (x[0] + y, x[1] + 1)
    except TypeError:
        return (x + y, 2)

truediv(*reduce(sum_and_count, L))

# prints 
20.11111111111111

Combining a couple of the above answers, I’ve come up with the following which works with reduce and doesn’t assume you have L available inside the reducing function:

from operator import truediv

L = [15, 18, 2, 36, 12, 78, 5, 6, 9]

def sum_and_count(x, y):
    try:
        return (x[0] + y, x[1] + 1)
    except TypeError:
        return (x + y, 2)

truediv(*reduce(sum_and_count, L))

# prints 
20.11111111111111

回答 20

我想添加另一种方法

import itertools,operator
list(itertools.accumulate(l,operator.add)).pop(-1) / len(l)

I want to add just another approach

import itertools,operator
list(itertools.accumulate(l,operator.add)).pop(-1) / len(l)

回答 21

numbers = [0,1,2,3]

numbers[0] = input("Please enter a number")

numbers[1] = input("Please enter a second number")

numbers[2] = input("Please enter a third number")

numbers[3] = input("Please enter a fourth number")

print (numbers)

print ("Finding the Avarage")

avarage = int(numbers[0]) + int(numbers[1]) + int(numbers[2]) + int(numbers [3]) / 4

print (avarage)
numbers = [0,1,2,3]

numbers[0] = input("Please enter a number")

numbers[1] = input("Please enter a second number")

numbers[2] = input("Please enter a third number")

numbers[3] = input("Please enter a fourth number")

print (numbers)

print ("Finding the Avarage")

avarage = int(numbers[0]) + int(numbers[1]) + int(numbers[2]) + int(numbers [3]) / 4

print (avarage)