Finding the average of a list

Question

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

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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      gt  gt  gt  l    15  18  2  36  12  78  5  6  9   gt  gt  gt  running average   reduce lambda aggr  elem   aggr 0    elem  aggr 1  1   l   0 0 0    gt  gt  gt  running average 0   181 0  9   gt  gt  gt  running average 0  running average 1  20 111111111111111

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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

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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

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l    15  18  2  36  12  78  5  6  9  sum l    len l

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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

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I want to add just another approach  import itertools operator list itertools accumulate l operator add   pop -1    len l

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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

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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

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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

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sum l    float len l   is the right answer  but just for completeness you can compute an average with a single reduce    gt  gt  gt  reduce lambda x  y  x   y   float len l    l  0  20 111111111111114   Note that this can result in a slight rounding error    gt  gt  gt  sum l    float len l   20 111111111111111

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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

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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

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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

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You can use numpy mean   l    15  18  2  36  12  78  5  6  9   import numpy as np print np mean l

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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

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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

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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

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l    15  18  2  36  12  78  5  6  9   l   map float l  print    2f    sum l  len l

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Or use pandas s Series mean method   pd Series sequence  mean     Demo    gt  gt  gt  import pandas as pd  gt  gt  gt  l    15  18  2  36  12  78  5  6  9   gt  gt  gt  pd Series l  mean   20 11111111111111  gt  gt  gt     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

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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

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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    gt  gt  gt  l    15  18  2  36  12  78  5  6  9   gt  gt  gt  print reduce lambda x  y  x   y  l    len l  20  gt  gt  gt  sum l  len l  20   Floating values   gt  gt  gt  print reduce lambda x  y  x   y  l    float len l   20 1111111111  gt  gt  gt  print sum l  float len l   20 1111111111    Andrew Clark was correct on his statement

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Instead of casting to float  you can add 0 0 to the sum   def avg l       return sum l  0 0    len l

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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

[python] Finding the average of a list

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