问题
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1:
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2:
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(l) / len(l)
回答3:
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4:
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5:
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6:
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7:
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8:
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9:
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10:
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11:
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12:
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13:
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14:
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15:
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16:
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17:
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
l = map(float,l)
print '%.2f' %(sum(l)/len(l))
回答18:
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19:
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20:
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21:
I want to add just another approach
import itertools,operator
list(itertools.accumulate(l,operator.add)).pop(-1) / len(l)
回答22:
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)
来源：https://stackoverflow.com/questions/9039961/finding-the-average-of-a-list