我有两个来自 map 函数的关键值:NY和其他。因此,我的密钥的输出为:NY 1或Other1。仅这两种情况。

我的 map 功能:

    #!/usr/bin/env python
    import sys
    import csv
    import string

    reader = csv.reader(sys.stdin, delimiter=',')
    for entry in reader:
        if len(entry) == 22:
            registration_state=entry[16]
            print('{0}\t{1}'.format(registration_state,int(1)))

现在,我需要使用reducer来处理 map 输出。我的减少:
#!/usr/bin/env python
import sys
import string


currentkey = None
ny = 0
other = 0
# 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, values = line.split('\t', 1)
    values = int(values)
#If we are still on the same key...
    if key == 'NY':
        ny = ny + 1
    #Otherwise, if this is a new key...
    else:
        #If this is a new key and not the first key we've seen
        other = other + 1


#Compute/output result for the last key
print('{0}\t{1}'.format('NY',ny))
print('{0}\t{1}'.format('Other',other))

通过这些,mapreduce将提供两个输出结果文件,每个文件都包含NY和Others输出。即,其中包含:NY 1248,其他4677;另一个:NY 0,其他1000。这是因为两个减法从 map 上拆分了输出,所以生成了两个结果,通过合并(合并)最终输出将是结果。

但是,我想将我的reduce或map函数更改为仅对一个键进行每个简化的过程,即一个reduce仅将NY作为键值,而另一个对Other处理。我希望得到类似以下结果的结果:
NY 1258, Others 0; Another: NY 0, Others 5677.

如何调整功能以达到预期效果?

最佳答案

可能您需要使用Python迭代器和生成器。
这个link就是一个很好的例子。我尝试用相同的代码重新编写代码(未经测试)

映射器:

#!/usr/bin/env python
"""A more advanced Mapper, using Python iterators and generators."""

import sys

def main(separator='\t'):
    reader = csv.reader(sys.stdin, delimiter=',')
    for entry in reader:
    if len(entry) == 22:
        registration_state=entry[16]
        print '%s%s%d' % (registration_state, separator, 1)

if __name__ == "__main__":
    main()

reducer :
!/usr/bin/env python
"""A more advanced Reducer, using Python iterators and generators."""

from itertools import groupby
from operator import itemgetter
import sys

def read_mapper_output(file, separator='\t'):
    for line in file:
        yield line.rstrip().split(separator, 1)

def main(separator='\t'):
    for current_word, group in groupby(data, itemgetter(0)):
        try:
            total_count = sum(int(count) for current_word, count in group)
            print "%s%s%d" % (current_word, separator, total_count)
        except ValueError:
            # count was not a number, so silently discard this item
            pass

if __name__ == "__main__":
    main()

关于python - 如何通过识别python Hadoop中的键来处理Mapreduce,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/49085928/

10-16 22:38