Finding the average of a list
PythonListLambdaAverageReducePython Problem Overview
I have to find the average of a list in Python. This is my code so far
from functools import reduce
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?
Python Solutions
Solution 1 - Python
On Python 3.8+, with floats, you can use statistics.fmean as it's faster with floats.
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:
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 functools.reduce as it is much slower.
Solution 2 - Python
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(l) / len(l)
Solution 3 - Python
You can use numpy.mean:
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
import numpy as np
print(np.mean(l))
Solution 4 - Python
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)
Solution 5 - Python
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 in Python 2 to force Python to do a floating-point division.)
Solution 6 - Python
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.
Solution 7 - Python
Instead of casting to float, you can add 0.0 to the sum:
def avg(l):
return sum(l, 0.0) / len(l)
Solution 8 - Python
EDIT:
I added two other ways to get the average of a list (which are relevant only for Python 3.8+). Here is the comparison that I made:
# test mean caculation
import timeit
import statistics
import numpy as np
from functools import reduce
import pandas as pd
import math
LIST_RANGE = 10000000000
NUMBERS_OF_TIMES_TO_TEST = 10000
l = list(range(10))
def mean1():
return statistics.mean(l)
def mean2():
return sum(l) / len(l)
def mean3():
return np.mean(l)
def mean4():
return np.array(l).mean()
def mean5():
return reduce(lambda x, y: x + y / float(len(l)), l, 0)
def mean6():
return pd.Series(l).mean()
def mean7():
return statistics.fmean(l)
def mean8():
return math.fsum(l) / len(l)
for func in [mean1, mean2, mean3, mean4, mean5, mean6, mean7, mean8 ]:
print(f"{func.__name__} took: ", timeit.timeit(stmt=func, number=NUMBERS_OF_TIMES_TO_TEST))
These are the results I got:
mean1 took: 0.09751558300000002
mean2 took: 0.005496791999999973
mean3 took: 0.07754683299999998
mean4 took: 0.055743208000000044
mean5 took: 0.018134082999999968
mean6 took: 0.6663848750000001
mean7 took: 0.004305374999999945
mean8 took: 0.003203333000000086
Interesting! looks like math.fsum(l) / len(l) is the fastest way, then statistics.fmean(l), and only then sum(l) / len(l). Nice!
Thank you @Asclepius for showing me these two other ways!
OLD ANSWER:
In terms of efficiency and speed, these are the results that I got testing the other answers:
# test mean caculation
import timeit
import statistics
import numpy as np
from functools import reduce
import pandas as pd
LIST_RANGE = 10000000000
NUMBERS_OF_TIMES_TO_TEST = 10000
l = list(range(10))
def mean1():
return statistics.mean(l)
def mean2():
return sum(l) / len(l)
def mean3():
return np.mean(l)
def mean4():
return np.array(l).mean()
def mean5():
return reduce(lambda x, y: x + y / float(len(l)), l, 0)
def mean6():
return pd.Series(l).mean()
for func in [mean1, mean2, mean3, mean4, mean5, mean6]:
print(f"{func.__name__} took: ", timeit.timeit(stmt=func, number=NUMBERS_OF_TIMES_TO_TEST))
and the results:
mean1 took: 0.17030245899968577
mean2 took: 0.002183011999932205
mean3 took: 0.09744236000005913
mean4 took: 0.07070840100004716
mean5 took: 0.022754742999950395
mean6 took: 1.6689282460001778
so clearly the winner is:
sum(l) / len(l)
Solution 9 - Python
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
Solution 10 - Python
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
Solution 11 - Python
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:
Solution 12 - Python
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.
Solution 13 - Python
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)
Solution 14 - Python
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)
Solution 15 - Python
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
Solution 16 - Python
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.
Solution 17 - Python
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
Solution 18 - Python
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
l = map(float,l)
print '%.2f' %(sum(l)/len(l))
Solution 19 - Python
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.
Solution 20 - Python
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
Solution 21 - Python
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
Solution 22 - Python
I want to add just another approach
import itertools,operator
list(itertools.accumulate(l,operator.add)).pop(-1) / len(l)
Solution 23 - Python
You can make a function for averages, usage:
average(21,343,2983) # You can pass as many arguments as you want.
Here is the code:
def average(*args):
total = 0
for num in args:
total+=num
return total/len(args)
*args allows for any number of answers.
Solution 24 - Python
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)