How to Calculate Standard Deviation in Python

Here’s how to calculate standard deviation in Python: for a plain list, use statistics.stdev() for a sample or statistics.pstdev() for a whole population. With arrays, use numpy.std(), but watch its default:

import statistics
statistics.stdev([1250.50, 980.25, 1430.00])    # sample standard deviation

import numpy as np
np.std(data, ddof=1)                             # sample; plain np.std is population

I calculated everything below on Python 3.12.5, NumPy 2.5.3 and pandas 3.0.6 using a week of made-up store sales in USD.

You’ll get the built-in statistics module, the formula by hand, the NumPy and pandas difference that trips everyone up, standard deviation by group, and using it to flag outliers.

Standard deviation with the statistics module

You don’t need to install anything for lists; the standard library has it covered. stdev() treats the data as a sample and divides by n – 1; pstdev() treats it as the entire population and divides by n:

import statistics

sales = [1250.50, 980.25, 1430.00, 1105.75, 760.00, 1320.40, 1188.90]   # one week, USD

print(f"mean:               ${statistics.mean(sales):,.2f}")
print(f"sample stdev:       ${statistics.stdev(sales):,.2f}")     # divides by n - 1
print(f"population pstdev:  ${statistics.pstdev(sales):,.2f}")    # divides by n
print(f"variance (sample):  {statistics.variance(sales):,.2f}")

try:
    statistics.stdev([99.0])
except statistics.StatisticsError as err:
    print("StatisticsError:", err)

Output:

mean:               $1,147.97
sample stdev:       $224.31
population pstdev:  $207.67
variance (sample):  50,315.66
StatisticsError: stdev requires at least two data points

Use the sample version when your data’s a subset, which it usually is. For the average itself, see the mean function in Python.

Calculating standard deviation by hand

The formula is short: find the mean, square each value’s distance from it, average those squares, and take the square root. The only choice is dividing by n or n – 1:

sales = [1250.50, 980.25, 1430.00, 1105.75, 760.00, 1320.40, 1188.90]   # one week, USD

import math

n = len(sales)
mean = sum(sales) / n
squared = [(x - mean) ** 2 for x in sales]

sample_sd = math.sqrt(sum(squared) / (n - 1))
population_sd = math.sqrt(sum(squared) / n)
print(f"mean ${mean:,.2f} | sample ${sample_sd:,.2f} | population ${population_sd:,.2f}")

Output:

mean $1,147.97 | sample $224.31 | population $207.67

Seeing the steps makes the n versus n – 1 choice concrete, and it’s what the libraries do for you. Dividing by n – 1 (Bessel’s correction) slightly widens the estimate to account for using a sample mean.

Why do NumPy and pandas give different standard deviations?

This catches a lot of people. np.std() defaults to the population formula (ddof=0), while pandas .std() defaults to the sample formula (ddof=1):

import numpy as np
import pandas as pd
import statistics

sales = [1250.50, 980.25, 1430.00, 1105.75, 760.00, 1320.40, 1188.90]   # one week, USD

arr = np.array(sales)
print(f"np.std(arr):          {np.std(arr):.2f}   <- population (ddof=0) by default")
print(f"np.std(arr, ddof=1):  {np.std(arr, ddof=1):.2f}")
print(f"pd.Series.std():      {pd.Series(sales).std():.2f}   <- sample (ddof=1) by default")
print(f"statistics.stdev():   {statistics.stdev(sales):.2f}")

Output:

np.std(arr):          207.67   <- population (ddof=0) by default
np.std(arr, ddof=1):  224.31
pd.Series.std():      224.31   <- sample (ddof=1) by default
statistics.stdev():   224.31

ddof stands for delta degrees of freedom: the divisor is n – ddof. I always pass it explicitly so nobody has to remember the defaults.

Standard deviation by group in pandas

With a DataFrame, groupby().agg() gives the mean and standard deviation per group in one step. The coefficient of variation (std divided by mean) makes groups of different sizes comparable:

import pandas as pd

df = pd.DataFrame({
    "store": ["Austin"] * 5 + ["Denver"] * 5,
    "sales": [1250.50, 1310.00, 1195.25, 1280.75, 1240.00,
              620.00, 1850.50, 940.25, 1500.00, 780.00],
})

summary = df.groupby("store")["sales"].agg(["mean", "std"]).round(2)
summary["cv"] = (summary["std"] / summary["mean"]).round(3)    # spread relative to the mean
print(summary)

Output:

           mean     std     cv
store                         
Austin  1255.30   43.31  0.035
Denver  1138.15  518.24  0.455

For more on selecting the data first, see loc vs iloc in pandas.

Using standard deviation to find outliers

A z-score says how many standard deviations a value sits from the mean. Values beyond about 2 or 3 are worth your closer look:

import numpy as np

daily = np.array([1250.50, 980.25, 1430.00, 1105.75, 760.00, 1320.40, 1188.90,
                  1215.00, 3980.00, 1099.99, 1275.25, 1160.60])
mean, sd = daily.mean(), daily.std(ddof=1)
z = (daily - mean) / sd

print(f"mean ${mean:,.2f}, sd ${sd:,.2f}")
for amount, score in zip(daily, z):
    flag = "  <- outlier" if abs(score) > 2 else ""
    print(f"${amount:>9,.2f}  z = {score:+.2f}{flag}")

Output:

mean $1,397.22, sd $831.21
$ 1,250.50  z = -0.18
$   980.25  z = -0.50
$ 1,430.00  z = +0.04
$ 1,105.75  z = -0.35
$   760.00  z = -0.77
$ 1,320.40  z = -0.09
$ 1,188.90  z = -0.25
$ 1,215.00  z = -0.22
$ 3,980.00  z = +3.11  <- outlier
$ 1,099.99  z = -0.36
$ 1,275.25  z = -0.15
$ 1,160.60  z = -0.28

One big outlier inflates the standard deviation itself, so for messy data you’ll want the median and interquartile range as a cross-check.

Which standard deviation function to use

Function	Default	Best for
statistics.stdev()	Sample (n – 1)	Lists, no extra installs
statistics.pstdev()	Population (n)	When you have every value
np.std(a, ddof=1)	Population unless ddof=1	Large arrays, speed
Series.std()	Sample (ddof=1)	DataFrames and groups

Continue with these stats and data guides:

• Average of a list in Python
• NumPy average
• Monte Carlo simulation in Python
• sum() function in Python

Frequently asked questions