# Pearsons Correlation Coefficient

Based on this StackOverflow answer by cbare.

## Preliminaries

```
import statistics as stats
```

## Create Data

```
x = [1,2,3,4,5,6,7,8,9]
y = [2,1,2,4.5,7,6.5,6,9,9.5]
```

## Calculate Pearson’s Correlation Coefficient

There are a number of equivalent expression ways to calculate Pearson’s correlation coefficient (also called Pearson’s r). Here is one.

$$r={\frac {1}{n-1}}\sum_{i=1}^{n}\left({\frac {x_{i}-{\bar {x}}}{s_{x}}}\right)\left({\frac {y_{i}-{\bar {y}}}{s_{y}}}\right)$$

where $s_{x}$ and $s_{y}$ are the sample standard deviation for $x$ and $y$, and $\left({\frac {x_{i}-{\bar {x}}}{s_{x}}}\right)$ is the standard score for $x$ and $y$.

```
# Create a function
def pearson(x,y):
# Create n, the number of observations in the data
n = len(x)
# Create lists to store the standard scores
standard_score_x = []
standard_score_y = []
# Calculate the mean of x
mean_x = stats.mean(x)
# Calculate the standard deviation of x
standard_deviation_x = stats.stdev(x)
# Calculate the mean of y
mean_y = stats.mean(y)
# Calculate the standard deviation of y
standard_deviation_y = stats.stdev(y)
# For each observation in x
for observation in x:
# Calculate the standard score of x
standard_score_x.append((observation - mean_x)/standard_deviation_x)
# For each observation in y
for observation in y:
# Calculate the standard score of y
standard_score_y.append((observation - mean_y)/standard_deviation_y)
# Multiple the standard scores together, sum them, then divide by n-1, return that value
return (sum([i*j for i,j in zip(standard_score_x, standard_score_y)]))/(n-1)
```

```
# Show Pearson's Correlation Coefficient
pearson(x,y)
```

```
0.9412443251336238
```