Question

非常愚蠢的问题.. 我有以下数据

id1, value
1, 20.2
1,20.4
....

我想找到id1的平均值和中位数？（注意..均值，每个id的中位数而不是全局均值，中位数）我正在使用python hadoop流..

mapper.py
for line in sys.stdin:
    try:
    # remove leading and trailing whitespace
        line = line.rstrip(os.linesep)
        tokens = line.split(",")

            print '%s,%s' % (tokens[0],tokens[1])
    except Exception:
        continue


reducer.py
data_dict = defaultdict(list)
def mean(data_list):
    return sum(data_list)/float(len(data_list)) if len(data_list) else 0
def median(mylist):
    sorts = sorted(mylist)
    length = len(sorts)
    if not length % 2:
        return (sorts[length / 2] + sorts[length / 2 - 1]) / 2.0
    return sorts[length / 2]


for line in sys.stdin:
    try:
        line = line.rstrip(os.linesep)
        serial_id, duration = line.split(",")
        data_dict[serial_id].append(float(duration))
    except Exception:
        pass
for k,v in data_dict.items():
    print "%s,%s,%s" %(k, mean(v), median(v))

我期待一个单一的平均值，每个键的中位数但是我看到id1重复了不同的均值和中位数.. 例如..做grep ..

mean_median/part-00003:SH002616940000,5.0,5.0   
mean_median/part-00008:SH002616940000,901.0,901.0   
mean_median/part-00018:SH002616940000,11.0,11.0 
mean_median/part-00000:SH002616940000,2.0,2.0   
mean_median/part-00025:SH002616940000,1800.0,1800.0 
mean_median/part-00002:SH002616940000,4.0,4.0   
mean_median/part-00006:SH002616940000,8.0,8.0   
mean_median/part-00021:SH002616940000,14.0,14.0 
mean_median/part-00001:SH002616940000,3.0,3.0   
mean_median/part-00022:SH002616940000,524.666666667,26.0    
mean_median/part-00017:SH002616940000,65.0,65.0 
mean_median/part-00016:SH002616940000,1384.0,1384.0 
mean_median/part-00020:SH002616940000,596.0,68.0    
mean_median/part-00014:SH002616940000,51.0,51.0 
mean_median/part-00004:SH002616940000,6.0,6.0   
mean_median/part-00005:SH002616940000,7.0,7.0

有什么建议吗？

Answer 1

我在hadoop-user邮件列表中回答了同样的问题如下：

你为这份工作开始了多少减速机？如果您为此作业启动了许多Reducers，它将生成多个输出文件，命名为part- * 。每个部分只是特定Reducer分区的局部均值和中值。

两种解决方案： 1，调用setNumReduceTasks（1）的方法将Reducer数设置为1，它只产生一个输出文件，每个不同的密钥只产生一个均值和中值。 2，在Hadoop源代码中引用org.apache.hadoop.examples.WordMedian。它通过本地函数处理由多个Reducer生成的所有输出文件，并产生最终结果。

使用python hadoop流式搜索平均中值

1 个答案: