Question

我有一个redis数据库，它每十秒钟从Arduino接收一次数据。

现在，我想制作六个十秒钟的数据来计算一个六十二秒的数据，然后获得六个十秒钟数据的平均值，最大值，最小值，如下所示。

import json

a = [u'{"id":"proximity_sensor1","tstamp":1440643570238,"avg":15.0,"coefVariation":0.0,"anom":0,"max":15.0,"min":15.0,"sample_size":10}',
 u'{"id":"proximity_sensor1","tstamp":1440643580307,"avg":15.0,"coefVariation":0.0,"anom":0,"max":15.0,"min":15.0,"sample_size":9}',
 u'{"id":"proximity_sensor1","tstamp":1440643590242,"avg":15.0,"coefVariation":0.0,"anom":0,"max":15.0,"min":15.0,"sample_size":9}',
 u'{"id":"proximity_sensor1","tstamp":1440643590242,"avg":15.0,"coefVariation":0.0,"anom":0,"max":15.0,"min":15.0,"sample_size":8}',
 u'{"id":"proximity_sensor1","tstamp":1440643590242,"avg":15.0,"coefVariation":0.0,"anom":0,"max":15.0,"min":15.0,"sample_size":9}',
 u'{"id":"proximity_sensor1","tstamp":1440643590242,"avg":15.0,"coefVariation":0.0,"anom":0,"max":15.0,"min":15.0,"sample_size":9}']

a = map(lambda x: json.loads(x), a)

#print a

def abc(aaa):
    for index in range(0, len(aaa), 6):
        abc = aaa[index:(index+6)]
        tuples = [('avg', 'max', 'min')]
        avg = sum(map(lambda x: x['avg'], abc))/6
        min_ = min(map(lambda x: x['min'], abc))
        max_ = max(map(lambda x: x['max'], abc))
        yield [avg, max_, min_]

print list(abc(a))

我在想是否有更好的解决方法。如果我使用itertools.groupby，我可以更快地解决它吗？或者任何人都有好主意简化计算过程？

Answer 1

正常情况下，itertools.groupby与某些条件一起使用来对元素进行分组，但是因为在你的情况下你没有任何这样的条件，相反你只想将每6个元素组合在一起，我不认为使用itertools .groupby会带来任何好处。

话虽如此，我可以提出一些其他改进建议 -

您可以使用＆＃39;键＆＃39; max / min函数的参数而不是map / lambda的当前方法，例如 -
```
max_ = max(abc, key= lambda x:x['max'])['max']
```
同样适用于min（）函数。
另外，我认为对sum（）而不是map / lambda进行列表理解会更具可读性。示例 -
```
 avg = sum([x['avg'] for x in abc])/6
```

我可以使用groupby在python中解决这个问题吗？

1 个答案: