python - PySpark 减少按键？添加键/元组

Question

我有以下数据，我要做的是

[(13, 'D'), (14, 'T'), (32, '6'), (45, 'T'), (47, '2'), (48, '0'), (49, '2'), (50, '0'), (51, 'T'), (53, '2'), (54, '0'), (13, 'A'), (14, 'T'), (32, '6'), (45, 'A'), (47, '2'), (48, '0'), (49, '2'), (50, '0'), (51, 'X')]

是为每个键计算值的实例(一个 1 字符串字符)。所以我先做了一张 map :

.map(lambda x: (x[0], [x[1], 1]))

现在将其作为键/元组:

[(13, ['D', 1]), (14, ['T', 1]), (32, ['6', 1]), (45, ['T', 1]), (47, ['2', 1]), (48, ['0', 1]), (49, ['2', 1]), (50, ['0', 1]), (51, ['T', 1]), (53, ['2', 1]), (54, ['0', 1]), (13, ['A', 1]), (14, ['T', 1]), (32, ['6', 1]), (45, ['A', 1]), (47, ['2', 1]), (48, ['0', 1]), (49, ['2', 1]), (50, ['0', 1]), (51, ['X', 1])]

我只是无法在最后一部分弄清楚如何为每个键计算该字母的实例。例如键 13 将有 1 个 D 和 1 个 A。而 14 将有 2 个 T，等等。

Answer 1

我更熟悉 Scala 中的 Spark，因此可能有比 Counter 更好的方法来计算 groupByKey 生成的可迭代对象中的字符数，但这里有一个选项:

from collections import Counter

rdd = sc.parallelize([(13, 'D'), (14, 'T'), (32, '6'), (45, 'T'), (47, '2'), (48, '0'), (49, '2'), (50, '0'), (51, 'T'), (53, '2'), (54, '0'), (13, 'A'), (14, 'T'), (32, '6'), (45, 'A'), (47, '2'), (48, '0'), (49, '2'), (50, '0'), (51, 'X')]) 
rdd.groupByKey().mapValues(Counter).collect()

[(48, Counter({'0': 2})),
 (32, Counter({'6': 2})),
 (49, Counter({'2': 2})),
 (50, Counter({'0': 2})),
 (51, Counter({'X': 1, 'T': 1})),
 (53, Counter({'2': 1})),
 (13, Counter({'A': 1, 'D': 1})),
 (45, Counter({'A': 1, 'T': 1})),
 (14, Counter({'T': 2})),
 (54, Counter({'0': 1})),
 (47, Counter({'2': 2}))]