我正在进行机器学习的在线课程。为了获得更好的预测，我们应该实现一个python函数来删除10％的异常数据。我已经解决了这个问题 - 但我找到了两种不同的解决方案，我希望根据代码质量听取您的意见。
def outlierCleaner(predictions, ages, net_worths):
"""
Clean away the 10% of points that have the largest
residual errors (difference between the prediction
and the actual net worth).
Return a list of tuples named cleaned_data where
each tuple is of the form (age, net_worth, error).
"""
# first approach just using list comprehension
import timeit
start = timeit.default_timer()
cleaned_data = []
cleaned_data = [(age, net_worth, (prediction-net_worth)**2)\
for age, net_worth, prediction \
in zip(ages, net_worths, predictions)]
last_index = int(len(cleaned_data)*0.9-1)
cleaned_data = sorted(cleaned_data, key = lambda t: t[2])
cleaned_data = cleaned_data[0:last_index]
stop = timeit.default_timer()
runtime1 = stop - start
# second approach. converting lists to numpy arrays, doing the math and
# converting back to a list of tuples.
start = timeit.default_timer()
import numpy as np
predictions_array = np.reshape(np.array(predictions), (len(predictions), 1))
ages_array = np.reshape(np.array(ages), (len(ages), 1))
net_worths_array = np.reshape(np.array(net_worths), (len(net_worths), 1))
cleaned_data = np.concatenate((ages_array, net_worths_array, (predictions_array-net_worths_array)**2), axis=1)
cleaned_data = cleaned_data[cleaned_data[:,2].argsort()]
cleaned_data = cleaned_data[:last_index+1, :]
#print "cleaned_data size:", cleaned_data.size, "dimens: ", cleaned_data.shape
cleaned_data = [tuple(row) for row in cleaned_data]
#print "length of cleaned_data:", len(cleaned_data)
runtime2 = stop - start
#print "runtime with lists: ", runtime1
#print "runtime with np.arrays: ", runtime2
#
# The second approach with numpy arrays seems to work a bit faster.
return cleaned_data
在numpy数组的帮助下，第二种方法似乎在我的机器上运行得更快（0.000577926635742与0.0）。
除了运行时这个版本，您认为哪个版本更合适？