Using numpy
and list comprehension
,
import numpy as np
data = [['6.5', '3.2', '5.1', '2.0', 'Iris-virginica'],
['6.1', '2.8', '4.0', '1.3', 'Iris-versicolor'] ,
['4.6', '3.2', '1.4', '0.2', 'Iris-setosa'],
['6.0', '2.2', '4.0', '1.0', 'Iris-versicolor'],
['4.7', '3.2', '1.3', '0.2', 'Iris-setosa'],
['6.7', '3.1', '5.6', '2.4', 'Iris-virginica']]
filtered = [map(float, item[:4]) for item in data if item[4] == 'Iris-virginica']
print 'mean', np.mean(filtered, axis=0)
print 'var ', np.var(filtered, axis=0)
where item[4] == 'Iris-virginica'
filters what you want, and map(float, item[:3])
is for str
to float
, then np.mean(..., axis=0)
is to get mean
of the filtered data.
The output is
mean [ 6.6 3.15 5.35]
var [ 0.01 0.0025 0.0625]
UPDATE
Here is numpy
only version, but this seems like slower than the above.
data = np.array(data)
filtered = data[data[:, 4] == 'Iris-virginica'][:, :3].astype(np.float)
print 'mean', np.mean(filtered, axis=0)
print 'var ', np.var(filtered, axis=0)
The timeit
result is
In [5]: %timeit filtered = [map(float, item[:4]) for item in data if item[4] == 'Iris-virginica']
100000 loops, best of 3: 1.93 µs per loop
In [6]: data = np.array(data)
In [7]: timeit data[data[:, 4] == 'Iris-virginica'][:, :4].astype(np.float)
100000 loops, best of 3: 15.5 µs per loop