Question

我有一个列表列表，需要根据列表中的字符串进行合并以适合结构。在这种情况下，它将是＆＃39; date＆＃39;并且＆＃39; id＆＃39;试图适应＆＃39;领域＆＃39;结构体。

字段：['date', 'id', 'impressions', 'clicks']

在：

[('2015-11-01', 'id123', 'impressions', '8'), ('2015-11-01', 'id123', 
'clicks', '4'), ('2015-11-01', 'id456', 'impressions', '14'), 
('2015-11-01', 'id456', 'clicks', '9')]

后：

[('2015-11-01', 'id123', '8', '4'), ('2015-11-01', 'id456', '14', '9')]

Answer 1

>>> L  = [('2015-11-01', 'id123', 'impressions', '8'), ('2015-11-01', 'id123', 
... 'clicks', '4'), ('2015-11-01', 'id456', 'impressions', '14'), 
... ('2015-11-01', 'id456', 'clicks', '9')]
>>> from collections import defaultdict
>>> D = defaultdict(list)
>>> for a, b, c, d in L:
...     D[a, b].append(d)
... 
>>> [k + tuple(D[k]) for k in D]
[('2015-11-01', 'id456', '14', '9'), ('2015-11-01', 'id123', '8', '4')]

如果展示次数和点击次数不一致

>>> L = [('2015-11-01', 'id123', 'impressions', '8'), ('2015-11-01', 'id123', 'clicks', '4'), ('2015-11-01', 'id456', 'clicks', '9'), ('2015-11-01', 'id456', 'impressions', '14')]
>>> from collections import defaultdict
>>> D = defaultdict(lambda: [None, None])
>>> for a, b, c, d in L:
...     D[a, b][c == 'clicks'] = d
... 
>>> [k + tuple(D[k]) for k in D]
[('2015-11-01', 'id456', '14', '9'), ('2015-11-01', 'id123', '8', '4')]

Answer 2

itertools.groupby在这里可以很好地工作，特别是如果真实数据与样本数据匹配（已经排序，因此日期/ id对都是相邻的）：

import itertools
from operator import itemgetter

outlist = []
for (date, ID), grp in itertools.groupby(inlist, key=itemgetter(0, 1)):
    grp = list(grp)  # Iterating twice, so convert to sequence
    impressioncnt = sum(int(cnt) for _, _, typ, cnt in grp if typ == 'impressions')
    clickcnt = sum(int(cnt) for _, _, typ, cnt in grp if typ == 'clicks')
    outlist.append((date, ID, str(impressioncnt), str(clickcnt)))

如果数据尚未按date和ID排序，则您需要先对inlist inlist.sort(key=itemgetter(0, 1))进行排序。如果list很大，那可能会很昂贵，在这种情况下，您可能会考虑使用collections.defaultdict代替：

import collections

dateID_cnts = collections.defaultdict({'impressions': 0, 'clicks': 0}.copy)
for date, ID, typ, cnt in inlist:
    dateID_cnts[date, ID][typ] += int(cnt)

# Convert from defaultdict to desired list of tuples
outlist = [(date, ID, str(v['impressions']), str(v['counts'])) for (date, ID), v in dateID_cnts.items()]

Answer 3

另一种方式：

data=[('2015-11-01', 'id123', 'impressions', '8'), 
      ('2015-11-01', 'id123','clicks', '4'), 
      ('2015-11-01', 'id456', 'impressions', '14'), 
      ('2015-11-01', 'id456', 'clicks', '9')]

ddict={}
for t in data:
    key=(t[0], t[1])
    ddict.setdefault(key, []).append(t[2:])

LoT=[]    
for d, id in ddict:
    impressions, clicks=max(ddict[(d, id)])[1], min(ddict[(d, id)])[1]
    LoT.append(tuple([d, id, impressions, clicks]))

>>> LoT
[('2015-11-01', 'id123', '8', '4'), ('2015-11-01', 'id456', '14', '9')]

如果您认为impressions和clicks已经有序，则可以删除max和min并将该行替换为：

impressions, clicks=ddict[(d, id)][0][1], ddict[(d, id)][1][1]

如何基于列表中的公共字符串Python来合并列表列表

3 个答案: