Question

我想应用map-reduce来处理python中的矩阵乘法与Hadoop。目标是计算A * B.输出应与输入类似。

输入是两个矩阵A和B格式，如下所示：

A,0,0,0.0
A,0,1,1.0
...
A,1,3,8.0
A,1,4,9.0
B,0,0,0.0
B,0,1,1.0
...
B,4,0,12.0
B,4,1,13.0

A，0,0,0.0表示索引为A（0,0），值为0.0，B表示相同。

这是我的地图功能：

import sys
import string
import numpy
#Split line into array of entry data
entry = line.split(",")
# Set row, column, and value for this entry
row = int(entry[1])
col = int(entry[2])
value = float(entry[3])

#If this is an entry in matrix A...
if (entry[0] == "A"):

        #Generate the necessary key-value pairs
        for i in range(col):
                print('<{}{},{} {} {}}>'.format(row,i,A,col,value))
#Otherwise, if this is an entry in matrix B...
else:
        #Generate the necessary key-value pairs
        for i in range(row):
                print('<{}{},{} {} {}}>'.format(i,col,B,row,value))

我想知道如何编写reduce函数。这是我将要使用的框架：

import sys
import string
import numpy

#number of columns of A/rows of B
n = int(sys.argv[1])

#Create data structures to hold the current row/column values (if needed; your code goes here)



currentkey = None

# input comes from STDIN (stream data that goes to the program)
for line in sys.stdin:

        #Remove leading and trailing whitespace
        line = line.strip()

        #Get key/value
        key, value = line.split('\t',1)

        #Parse key/value input (your code goes here)

    #If we are still on the same key...
    if key==currentkey:

            #Process key/value pair (your code goes here)


    #Otherwise, if this is a new key...
    else:
            #If this is a new key and not the first key we've seen
            if currentkey:

                    #compute/output result to STDOUT (your code goes here)

            currentkey = key

            #Process input for new key (your code goes here)

#Compute/output result for the last key (your code goes here)

要运行这两个函数，我将使用一个小测试数据集使用以下代码测试它们：

cat smalltest.txt | python src/map.py 2 3 | sort -n | python src/reduce.py 5

Map给出输出，然后使用sort -n对键进行排序，因此我将使用reducer来处理矩阵计算。我的困惑在于编写reducer函数。

Answer 1

不确定为什么减少
我的numpy方法（有一些字符串/列表/拉链体操）

 strin = '''A,0,0,0.0
A,0,1,1.0
A,1,0,8.0
A,1,1,9.0
B,0,0,0.0
B,0,1,1.0
B,1,0,12.0
B,1,1,13.0'''.split()

lines = [*map(lambda x: x.split(","),strin)]

linesT = [*zip(*lines)]

linesT

[('A', 'A', 'A', 'A', 'B', 'B', 'B', 'B'),
 ('0', '0', '1', '1', '0', '0', '1', '1'),
 ('0', '1', '0', '1', '0', '1', '0', '1'),
 ('0.0', '1.0', '8.0', '9.0', '0.0', '1.0', '12.0', '13.0')]

现在我们可以获得dims，数组A，B的数据

lastA = linesT[0].index("B") - 1

rowsA, colsA = int(linesT[1][lastA]) + 1, int(linesT[2][lastA]) + 1

datA = [*map(float, linesT[3][0:lastA + 1])]

A = np.array(datA).reshape((rowsA, colsA))

A
Out[50]: 
array([[ 0.,  1.],
       [ 8.,  9.]])

firstB = lastA + 1

rowsB, colsB = int(linesT[1][-1]) + 1, int(linesT[2][-1]) + 1

datB = [*map(float, linesT[3][firstB::])]

B = np.array(datB).reshape((rowsB, colsB))

B
Out[51]: 
array([[  0.,   1.],
       [ 12.,  13.]])

A @ B
Out[52]: 
array([[  12.,   13.],
       [ 108.,  125.]])

Answer 2

好吧，生病直截了当，

    lines = [*map(lambda x: x.split(","),strin)]

是简化的方法，如果lambda函数本身甚至不在带语法的输入中，那就好像字符串不存在一样减少它是老实说，你应该感谢，这段代码（不要太苛刻）是凌乱的，所以我不明白为什么你抱怨自动减少..

Map-Reduce用Hadoop解决python中的矩阵乘法

2 个答案: