Thomas Matthew - 11 months ago 98

Python Question

With the pandas dataframe below, taken from a dict of dict:

`import numpy as np`

import pandas as pd

from scipy.stats import pearsonr

NaN = np.nan

dd ={'A': {'A': '1', 'B': '2', 'C': '3'},

'B': {'A': '4', 'B': '5', 'C': '6'},

'C': {'A': '7', 'B': '8', 'C': '9'}}

df_link_link = pd.DataFrame.from_dict(dd, orient='index')

I'd like to form a new pandas DataFrame with the results of a Pearson correlation between rows for every row, excluding Pearson correlations between the same rows (correlating A with itself should just be

`NaN`

`dict_congruent = {'A': {'A': NaN,`

'B': pearsonr([NaN,2,3],[4,5,6]),

'C': pearsonr([NaN,2,3],[7,8,9])},

'B': {'A': pearsonr([4,NaN,6],[1,2,3]),

'B': NaN,

'C': pearsonr([4,NaN,6],[7,8,9])},

'C': {'A': pearsonr([7,8,NaN],[1,2,3]),

'B': pearsonr([7,8,NaN],[4,5,6]),

'C': NaN }}

where

`NaN`

`numpy.nan`

Answer Source

*Canonical but not viable solution*

```
df.corr().mask(np.equal.outer(df.index.values, df.columns.values))
```

default method for `corr`

is `pearson`

.

*TL;DR*

*Transpose Your Data To Use This*

*Wrapped up with a bow*

*very wide data*

```
np.random.seed([3,1415])
m, n = 1000, 10000
df = pd.DataFrame(np.random.randn(m, n), columns=['s{}'.format(i) for i in range(n)])
```

*magic function*

```
def corr(df, step=100, mask_diagonal=False):
def corr_closure(df):
d = df.values
sums = d.sum(0, keepdims=True)
stds = d.std(0, keepdims=True)
n = d.shape[0]
def corr_(k=0, l=10):
d2 = d.T.dot(d[:, k:l])
sums2 = sums.T.dot(sums[:, k:l])
stds2 = stds.T.dot(stds[:, k:l])
return pd.DataFrame((d2 - sums2 / n) / stds2 / n,
df.columns, df.columns[k:l])
return corr_
c = corr_closure(df)
step = min(step, df.shape[1])
tups = zip(range(0, n, step), range(step, n + step, step))
corr_table = pd.concat([c(*t) for t in tups], axis=1)
if mask_diagonal:
np.fill_diagonal(corr_table.values, np.nan)
return corr_table
```

*demonstration*

```
ct = corr(df, mask_diagonal=True)
ct.iloc[:10, :10]
```

*Magic Solution Explained*

*Logic:*

- use a closure to pre-calculate column sums and standard deviations
- return a function that takes positions of columns over which to correlate

```
def corr_closure(df):
d = df.values # get underlying numpy array
sums = d.sum(0, keepdims=True) # pre calculate sums
stds = d.std(0, keepdims=True) # pre calculate standard deviations
n = d.shape[0] # grab number of rows
def corr(k=0, l=10):
d2 = d.T.dot(d[:, k:l]) # for this slice, run dot product
sums2 = sums.T.dot(sums[:, k:l]) # dot pre computed sums with slice
stds2 = stds.T.dot(stds[:, k:l]) # dot pre computed stds with slice
# calculate correlations with the information I have
return pd.DataFrame((d2 - sums2 / n) / stds2 / n,
df.columns, df.columns[k:l])
return corr
```

*10000 columns*

`df.corr()`

did not finish in a reasonable amount of time