标签归档：average

Python NumPy中的np.mean（）vs np.average（）吗？

2021年8月2日 Python实用宝典

问题：Python NumPy中的np.mean（）vs np.average（）吗？

我注意到

In [30]: np.mean([1, 2, 3])
Out[30]: 2.0

In [31]: np.average([1, 2, 3])
Out[31]: 2.0

但是，应该存在一些差异，因为它们毕竟是两个不同的功能。

它们之间有什么区别？

I notice that

In [30]: np.mean([1, 2, 3])
Out[30]: 2.0

In [31]: np.average([1, 2, 3])
Out[31]: 2.0

However, there should be some differences, since after all they are two different functions.

What are the differences between them?

回答 0

np.average采用可选的权重参数。如果未提供，则等效。看一下源代码：Mean，Average

np.mean：

try:
    mean = a.mean
except AttributeError:
    return _wrapit(a, 'mean', axis, dtype, out)
return mean(axis, dtype, out)

np.average：

...
if weights is None :
    avg = a.mean(axis)
    scl = avg.dtype.type(a.size/avg.size)
else:
    #code that does weighted mean here

if returned: #returned is another optional argument
    scl = np.multiply(avg, 0) + scl
    return avg, scl
else:
    return avg
...

np.average takes an optional weight parameter. If it is not supplied they are equivalent. Take a look at the source code: Mean, Average

np.mean:

try:
    mean = a.mean
except AttributeError:
    return _wrapit(a, 'mean', axis, dtype, out)
return mean(axis, dtype, out)

np.average:

...
if weights is None :
    avg = a.mean(axis)
    scl = avg.dtype.type(a.size/avg.size)
else:
    #code that does weighted mean here

if returned: #returned is another optional argument
    scl = np.multiply(avg, 0) + scl
    return avg, scl
else:
    return avg
...

回答 1

np.mean 总是计算算术平均值，并具有一些用于输入和输出的其他选项（例如，使用什么数据类型，将结果放置在何处）。

np.average如果weights提供了参数，则可以计算加权平均值。

np.mean always computes an arithmetic mean, and has some additional options for input and output (e.g. what datatypes to use, where to place the result).

np.average can compute a weighted average if the weights parameter is supplied.

回答 2

在某些版本的numpy中，您必须意识到另一个重要的区别：

average 不考虑掩码，因此请计算整个数据集的平均值。

mean 考虑到掩码，因此仅对未掩码的值计算平均值。

g = [1,2,3,55,66,77]
f = np.ma.masked_greater(g,5)

np.average(f)
Out: 34.0

np.mean(f)
Out: 2.0

In some version of numpy there is another imporant difference that you must be aware:

average do not take in account masks, so compute the average over the whole set of data.

mean takes in account masks, so compute the mean only over unmasked values.

g = [1,2,3,55,66,77]
f = np.ma.masked_greater(g,5)

np.average(f)
Out: 34.0

np.mean(f)
Out: 2.0

回答 3

在您的调用中，两个函数是相同的。

average 可以计算加权平均值。

Doc链接：mean和average

In your invocation, the two functions are the same.

average can compute a weighted average though.

Doc links: mean and average

回答 4

除了已经指出的差异之外，还有另一个非常重要的差异，我刚刚发现了很难的方法：与不同np.mean，np.average不允许使用dtype关键字，这在某些情况下对于获得正确的结果至关重要。我有一个非常大的单精度数组，可以从h5文件访问它。如果我沿轴0和1取平均值，除非指定dtype='float64'：

>T.shape
(4096, 4096, 720)
>T.dtype
dtype('<f4')

m1 = np.average(T, axis=(0,1))                #  garbage
m2 = np.mean(T, axis=(0,1))                   #  the same garbage
m3 = np.mean(T, axis=(0,1), dtype='float64')  # correct results

不幸的是，除非您知道要查找的内容，否则不一定能说出结果是错误的。np.average由于这个原因，我将不再使用，但将始终np.mean(.., dtype='float64')在任何大型阵列上使用。如果我想要一个加权平均数，我将使用权重向量和目标数组的乘积，然后再加上np.sum或np.mean，适当地（也具有适当的精度），对它进行显式计算。

In addition to the differences already noted, there’s another extremely important difference that I just now discovered the hard way: unlike np.mean, np.average doesn’t allow the dtype keyword, which is essential for getting correct results in some cases. I have a very large single-precision array that is accessed from an h5 file. If I take the mean along axes 0 and 1, I get wildly incorrect results unless I specify dtype='float64':

>T.shape
(4096, 4096, 720)
>T.dtype
dtype('<f4')

m1 = np.average(T, axis=(0,1))                #  garbage
m2 = np.mean(T, axis=(0,1))                   #  the same garbage
m3 = np.mean(T, axis=(0,1), dtype='float64')  # correct results

Unfortunately, unless you know what to look for, you can’t necessarily tell your results are wrong. I will never use np.average again for this reason but will always use np.mean(.., dtype='float64') on any large array. If I want a weighted average, I’ll compute it explicitly using the product of the weight vector and the target array and then either np.sum or np.mean, as appropriate (with appropriate precision as well).

知识问答

在Python中计算算术平均值（一种平均值）

2021年7月31日 Python实用宝典

问题：在Python中计算算术平均值（一种平均值）

Python中是否有内置或标准库方法来计算数字列表的算术平均值（一种平均值）？

Is there a built-in or standard library method in Python to calculate the arithmetic mean (one type of average) of a list of numbers?

回答 0

我不知道标准库中的任何内容。但是，您可以使用类似以下内容的方法：

def mean(numbers):
    return float(sum(numbers)) / max(len(numbers), 1)

>>> mean([1,2,3,4])
2.5
>>> mean([])
0.0

在numpy中，有numpy.mean()。

I am not aware of anything in the standard library. However, you could use something like:

def mean(numbers):
    return float(sum(numbers)) / max(len(numbers), 1)

>>> mean([1,2,3,4])
2.5
>>> mean([])
0.0

In numpy, there’s numpy.mean().

回答 1

NumPy的a numpy.mean是算术平均值。用法很简单：

>>> import numpy
>>> a = [1, 2, 4]
>>> numpy.mean(a)
2.3333333333333335

NumPy has a numpy.mean which is an arithmetic mean. Usage is as simple as this:

>>> import numpy
>>> a = [1, 2, 4]
>>> numpy.mean(a)
2.3333333333333335

回答 2

用途statistics.mean：

import statistics
print(statistics.mean([1,2,4])) # 2.3333333333333335

从Python 3.4开始可用。对于3.1-3.3用户，该模块的旧版本可在PyPI上以的名称获得stats。只需更改statistics为即可stats。

Use statistics.mean:

import statistics
print(statistics.mean([1,2,4])) # 2.3333333333333335

It’s available since Python 3.4. For 3.1-3.3 users, an old version of the module is available on PyPI under the name stats. Just change statistics to stats.

回答 3

您甚至不需要麻木或肮脏的…

>>> a = [1, 2, 3, 4, 5, 6]
>>> print(sum(a) / len(a))
3

You don’t even need numpy or scipy…

>>> a = [1, 2, 3, 4, 5, 6]
>>> print(sum(a) / len(a))
3

回答 4

使用scipy：

import scipy;
a=[1,2,4];
print(scipy.mean(a));

Use scipy:

import scipy;
a=[1,2,4];
print(scipy.mean(a));

回答 5

除了强制浮动之外，您还可以执行以下操作

def mean(nums):
    return sum(nums, 0.0) / len(nums)

或使用lambda

mean = lambda nums: sum(nums, 0.0) / len(nums)

更新日期：2019-12-15

Python 3.8 在统计模块中添加了功能fmean。哪个更快，并且总是返回float。

将数据转换为浮点并计算算术平均值。

它的运行速度比mean（）函数快，并且始终返回浮点数。数据可以是序列或可迭代的。如果输入数据集为空，则引发StatisticsError。

fmean（[3.5，4.0，5.25]）

4.25

3.8版的新功能。

Instead of casting to float you can do following

def mean(nums):
    return sum(nums, 0.0) / len(nums)

or using lambda

mean = lambda nums: sum(nums, 0.0) / len(nums)

UPDATES: 2019-12-15

Python 3.8 added function fmean to statistics module. Which is faster and always returns float.

Convert data to floats and compute the arithmetic mean.

This runs faster than the mean() function and it always returns a float. The data may be a sequence or iterable. If the input dataset is empty, raises a StatisticsError.

fmean([3.5, 4.0, 5.25])

4.25

New in version 3.8.

回答 6

from statistics import mean
avarage=mean(your_list)

例如

from statistics import mean

my_list=[5,2,3,2]
avarage=mean(my_list)
print(avarage)

结果是

3.0

from statistics import mean
avarage=mean(your_list)

for example

from statistics import mean

my_list=[5,2,3,2]
avarage=mean(my_list)
print(avarage)

and result is

3.0

回答 7

def avg(l):
    """uses floating-point division."""
    return sum(l) / float(len(l))

例子：

l1 = [3,5,14,2,5,36,4,3]
l2 = [0,0,0]

print(avg(l1)) # 9.0
print(avg(l2)) # 0.0

def avg(l):
    """uses floating-point division."""
    return sum(l) / float(len(l))

Examples:

l1 = [3,5,14,2,5,36,4,3]
l2 = [0,0,0]

print(avg(l1)) # 9.0
print(avg(l2)) # 0.0

回答 8

def list_mean(nums):
    sumof = 0
    num_of = len(nums)
    mean = 0
    for i in nums:
        sumof += i
    mean = sumof / num_of
    return float(mean)

def list_mean(nums):
    sumof = 0
    num_of = len(nums)
    mean = 0
    for i in nums:
        sumof += i
    mean = sumof / num_of
    return float(mean)

回答 9

我一直认为应该avg从Builtins / stdlib中省略它，因为它很简单

sum(L)/len(L) # L is some list

并且任何告诫将在调用者代码中解决以供本地使用。

注意事项：

非浮点结果：在python2中，9/4为2。解析，使用float(sum(L))/len(L)或from __future__ import division

除以零：列表可能为空。解决：

if not L:
    raise WhateverYouWantError("foo")
avg = float(sum(L))/len(L)

I always supposed avg is omitted from the builtins/stdlib because it is as simple as

sum(L)/len(L) # L is some list

and any caveats would be addressed in caller code for local usage already.

Notable caveats:

non-float result: in python2, 9/4 is 2. to resolve, use float(sum(L))/len(L) or from __future__ import division

division by zero: the list may be empty. to resolve:

if not L:
    raise WhateverYouWantError("foo")
avg = float(sum(L))/len(L)

回答 10

对您问题的正确答案是使用statistics.mean。但是为了好玩，这是一个不使用该len()功能的Mean的版本，因此statistics.mean可以在不支持该功能的生成器上使用它（如）len()：

from functools import reduce
from operator import truediv
def ave(seq):
    return truediv(*reduce(lambda a, b: (a[0] + b[1], b[0]), 
                           enumerate(seq, start=1), 
                           (0, 0)))

The proper answer to your question is to use statistics.mean. But for fun, here is a version of mean that does not use the len() function, so it (like statistics.mean) can be used on generators, which do not support len():

from functools import reduce
from operator import truediv
def ave(seq):
    return truediv(*reduce(lambda a, b: (a[0] + b[1], b[0]), 
                           enumerate(seq, start=1), 
                           (0, 0)))

回答 11

其他人已经发布了很好的答案，但有些人可能仍在寻找找到Mean（avg）的经典方法，因此，我在这里发布了此信息（在Python 3.6中测试过的代码）：

def meanmanual(listt):

mean = 0
lsum = 0
lenoflist = len(listt)

for i in listt:
    lsum += i

mean = lsum / lenoflist
return float(mean)

a = [1, 2, 3, 4, 5, 6]
meanmanual(a)

Answer: 3.5

Others already posted very good answers, but some people might still be looking for a classic way to find Mean(avg), so here I post this (code tested in Python 3.6):

def meanmanual(listt):

mean = 0
lsum = 0
lenoflist = len(listt)

for i in listt:
    lsum += i

mean = lsum / lenoflist
return float(mean)

a = [1, 2, 3, 4, 5, 6]
meanmanual(a)

Answer: 3.5

知识问答

查找列表的平均值

2021年7月25日 Python实用宝典

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，并已在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pandas的Series.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)

Python 实用宝典

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