Apache Spark不如Scikit Learn准确吗？

Question

我最近一直试图了解Apache Spark作为Scikit Learn的替代品,但在我看来,即使在简单的情况下,Scikit也会比Spark更快地收敛到精确模型.例如,我使用以下脚本为非常简单的线性函数(z = x + y)生成了1000个数据点:

from random import random

def func(in_vals):
    '''result = x (+y+z+w....)'''
    result = 0
    for v in in_vals:
        result += v
    return result

if __name__ == "__main__":
    entry_count = 1000
    dim_count = 2
    in_vals = [0]*dim_count
    with open("data_yequalsx.csv", "w") as out_file:
        for entry in range(entry_count):
            for i in range(dim_count):
                in_vals[i] = random()
            out_val = func(in_vals)
            out_file.write(','.join([str(x) for x in in_vals]))
            out_file.write(",%s\n" % str(out_val))

然后我运行了以下Scikit脚本:

import sklearn
from sklearn import linear_model

import numpy as np

data = []
target = []
with open("data_yequalsx.csv") as inFile:
    for row in inFile:
        vals = row.split(",")
        data.append([float(x) for x in vals[:-1]])
        target.append(float(vals[-1]))

test_samples= len(data)/10

train_data = [0]*(len(data) - test_samples)
train_target = [0]*(len(data) - test_samples)
test_data = [0]*(test_samples)
test_target = [0]*(test_samples)
train_index = 0
test_index = 0
for j in range(len(data)):
    if j >= test_samples:
        train_data[train_index] = data[j]
        train_target[train_index] = target[j]
        train_index += 1
    else:
        test_data[test_index] = data[j]
        test_target[test_index] = target[j]
        test_index += 1

model = linear_model.SGDRegressor(n_iter=100, learning_rate="invscaling", eta0=0.0001, power_t=0.5, penalty="l2", alpha=0.0001, loss="squared_loss")
model.fit(train_data, train_target)
print(model.coef_)
print(model.intercept_)

result = model.predict(test_data)
mse = np.mean((result - test_target) ** 2)
print("Mean Squared Error = %s" % str(mse))

然后这个Spark脚本:(使用spark-submit,没有其​​他参数)

from pyspark.mllib.regression import LinearRegressionWithSGD, LabeledPoint
from pyspark import SparkContext

sc = SparkContext (appName="mllib_simple_accuracy")

raw_data = sc.textFile ("data_yequalsx.csv", minPartitions=10) #MinPartitions doesnt guarantee that you get that many partitions, just that you wont have fewer than that many partitions
data = raw_data.map(lambda line: [float(x) for x in line.split (",")]).map(lambda entry: LabeledPoint (entry[-1], entry[:-1])).zipWithIndex()
test_samples= data.count()/10

training_data = data.filter(lambda (entry, index): index >= test_samples).map(lambda (lp,index): lp)
test_data = data.filter(lambda (entry, index): index < test_samples).map(lambda (lp,index): lp)

model = LinearRegressionWithSGD.train(training_data, step=0.01, iterations=100, regType="l2", regParam=0.0001, intercept=True)
print(model._coeff)
print(model._intercept)

mse = (test_data.map(lambda lp: (lp.label - model.predict(lp.features))**2 ).reduce(lambda x,y: x+y))/test_samples;
print("Mean Squared Error: %s" % str(mse))

sc.stop ()

奇怪的是,火花给出的误差比Scikit给出的误差大一个数量级(分别为0.185和0.045),尽管两个模型具有几乎相同的设置(据我所知)我明白这是使用SGD与很少的迭代,所以结果可能会有所不同,但我不会认为它会接近这么大的差异或如此大的错误,特别是考虑到非常简单的数据.

---

我在Spark中有什么误解吗？它配置不正确吗？当然我应该得到一个比这更小的错误？

Answer 1

由于 Spark 是并行的，因此当计算正在进行时，每个节点都需要能够独立于其他节点工作，以避免节点之间昂贵的洗牌。因此，它使用称为随机梯度下降的过程来接近最小值，该最小值遵循向下的局部梯度。

解决[简单、最小二乘]回归问题的“精确”方法涉及求解矩阵方程。这可能就是 Scikit-Learn 正在做的事情，所以在这种情况下它会更准确。

需要权衡的是，对于大小为 N 的方阵，求解矩阵方程通常会缩放为 N^3，这对于大型数据集很快变得不可行。Spark 用准确性换取计算能力。与任何机器学习过程一样，您应该在整个算法中进行大量健全性检查，以确保上一步的结果有意义。

希望这可以帮助！