CODEWITHSUNDEEP
pythonliststandard-deviation
physics_for_all
physics_for_all

Reputation: 2263

Standard deviation of a list

I want to find mean and standard deviation of 1st, 2nd,... digits of several (Z) lists. For example, I have

A_rank=[0.8,0.4,1.2,3.7,2.6,5.8]
B_rank=[0.1,2.8,3.7,2.6,5,3.4]
C_Rank=[1.2,3.4,0.5,0.1,2.5,6.1]
# etc (up to Z_rank )...

Now I want to take the mean and std of *_Rank[0], the mean and std of *_Rank[1], etc.
(ie: mean and std of the 1st digit from all the (A..Z)_rank lists;
the mean and std of the 2nd digit from all the (A..Z)_rank lists;
the mean and std of the 3rd digit...; etc).

Upvotes: 154

Views: 438308

Answers (8)

pankaj
pankaj

Reputation: 1014

Using python, here are few methods:

import statistics as st

n = int(input())
data = list(map(int, input().split()))

Approach1 - using a function

stdev = st.pstdev(data)

Approach2: calculate variance and take square root of it

variance = st.pvariance(data)
devia = math.sqrt(variance)

Approach3: using basic math

mean = sum(data)/n
variance = sum([((x - mean) ** 2) for x in X]) / n
stddev = variance ** 0.5

print("{0:0.1f}".format(stddev))

Note:

  • variance calculates variance of sample population
  • pvariance calculates variance of entire population
  • similar differences between stdev and pstdev

Upvotes: 15

Alex Riley
Alex Riley

Reputation: 176810

Here's some pure-Python code you can use to calculate the mean and standard deviation.

All code below is based on the statistics module in Python 3.4+.

def mean(data):
    """Return the sample arithmetic mean of data."""
    n = len(data)
    if n < 1:
        raise ValueError('mean requires at least one data point')
    return sum(data)/n # in Python 2 use sum(data)/float(n)

def _ss(data):
    """Return sum of square deviations of sequence data."""
    c = mean(data)
    ss = sum((x-c)**2 for x in data)
    return ss

def stddev(data, ddof=0):
    """Calculates the population standard deviation
    by default; specify ddof=1 to compute the sample
    standard deviation."""
    n = len(data)
    if n < 2:
        raise ValueError('variance requires at least two data points')
    ss = _ss(data)
    pvar = ss/(n-ddof)
    return pvar**0.5

Note: for improved accuracy when summing floats, the statistics module uses a custom function _sum rather than the built-in sum which I've used in its place.

Now we have for example:

>>> mean([1, 2, 3])
2.0
>>> stddev([1, 2, 3]) # population standard deviation
0.816496580927726
>>> stddev([1, 2, 3], ddof=1) # sample standard deviation
0.1

Upvotes: 55

Bengt
Bengt

Reputation: 14448

Since Python 3.4 / PEP450 there is a statistics module in the standard library, which has a method stdev for calculating the standard deviation of iterables like yours:

>>> A_rank = [0.8, 0.4, 1.2, 3.7, 2.6, 5.8]
>>> import statistics
>>> statistics.stdev(A_rank)
2.0634114147853952

Upvotes: 206

Elad Yehezkel
Elad Yehezkel

Reputation: 111

pure python code:

from math import sqrt

def stddev(lst):
    mean = float(sum(lst)) / len(lst)
    return sqrt(float(reduce(lambda x, y: x + y, map(lambda x: (x - mean) ** 2, lst))) / len(lst))

Upvotes: 5

Samie Bee
Samie Bee

Reputation: 1365

The other answers cover how to do std dev in python sufficiently, but no one explains how to do the bizarre traversal you've described.

I'm going to assume A-Z is the entire population. If not see Ome's answer on how to inference from a sample.

So to get the standard deviation/mean of the first digit of every list you would need something like this:

#standard deviation
numpy.std([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

#mean
numpy.mean([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

To shorten the code and generalize this to any nth digit use the following function I generated for you:

def getAllNthRanks(n):
    return [A_rank[n], B_rank[n], C_rank[n], D_rank[n], E_rank[n], F_rank[n], G_rank[n], H_rank[n], I_rank[n], J_rank[n], K_rank[n], L_rank[n], M_rank[n], N_rank[n], O_rank[n], P_rank[n], Q_rank[n], R_rank[n], S_rank[n], T_rank[n], U_rank[n], V_rank[n], W_rank[n], X_rank[n], Y_rank[n], Z_rank[n]]

Now you can simply get the stdd and mean of all the nth places from A-Z like this:

#standard deviation
numpy.std(getAllNthRanks(n))

#mean
numpy.mean(getAllNthRanks(n))

Upvotes: 3

Ome
Ome

Reputation: 331

In Python 2.7.1, you may calculate standard deviation using numpy.std() for:

  • Population std: Just use numpy.std() with no additional arguments besides to your data list.
  • Sample std: You need to pass ddof (i.e. Delta Degrees of Freedom) set to 1, as in the following example:

numpy.std(< your-list >, ddof=1)

The divisor used in calculations is N - ddof, where N represents the number of elements. By default ddof is zero.

It calculates sample std rather than population std.

Upvotes: 22

B.Kocis
B.Kocis

Reputation: 2020

In python 2.7 you can use NumPy's numpy.std() gives the population standard deviation.

In Python 3.4 statistics.stdev() returns the sample standard deviation. The pstdv() function is the same as numpy.std().

Upvotes: 13

NPE
NPE

Reputation: 500357

I would put A_Rank et al into a 2D NumPy array, and then use numpy.mean() and numpy.std() to compute the means and the standard deviations:

In [17]: import numpy

In [18]: arr = numpy.array([A_rank, B_rank, C_rank])

In [20]: numpy.mean(arr, axis=0)
Out[20]: 
array([ 0.7       ,  2.2       ,  1.8       ,  2.13333333,  3.36666667,
        5.1       ])

In [21]: numpy.std(arr, axis=0)
Out[21]: 
array([ 0.45460606,  1.29614814,  1.37355985,  1.50628314,  1.15566239,
        1.2083046 ])

Upvotes: 120

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