在Python中，如何将一个列表拆分为包含重复值的子列表？

Question

我有一个每天特定时间的潮汐信息列表。有点像这样：

tideData = [
['Thursday 4 January',11.58,0.38],
['Thursday 4 January',16.95,0.73],
['Friday 5 January',6.48,0.83],
['Friday 5 January',12.42,0.33],
['Saturday 6 January',0.5,0.02],
['Saturday 6 January',7.18,0.85],
...
['Friday 2 February',23.52,0.04]
]

我想将此列表拆分为包含相同日期的子列表。在上述情况下，列表将变为：

tideData = [
[['Thursday 4 January',11.58,0.38],
['Thursday 4 January',16.95,0.73]],
[['Friday 5 January',6.48,0.83],
['Friday 5 January',12.42,0.33],
['Friday 5 January',17.92,0.75]],
[['Saturday 6 January',0.5,0.02],
['Saturday 6 January',7.18,0.85]],
...
['Friday 2 February',23.52,0.04]]
]

现在，如果每次约会的次数相等，这就不是问题了。然而，日期有时出现两次，有时出现三次。因此，我希望能够根据重复日期将它们分类为子列表。我该怎么办？你知道吗

Answer 1

我想您应该使用groupby包中的itertools

from itertools import groupby

tideData = [
['Thursday 4 January',11.58,0.38],
['Thursday 4 January',16.95,0.73],
['Friday 5 January',6.48,0.83],
['Friday 5 January',12.42,0.33],
['Saturday 6 January',0.5,0.02],
['Saturday 6 January',7.18,0.85],
['Friday 2 February',23.52,0.04]
]

如果未对数据进行排序，则可以使用：

tideData = sorted(tideData, key=lambda x: x[0])

在使用以下工具之前：

[list(g) for _,g in groupby(tideData, key=lambda x: x[0])]
# returns:
[[['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73]],
 [['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]],
 [['Saturday 6 January', 0.5, 0.02], ['Saturday 6 January', 7.18, 0.85]],
 [['Friday 2 February', 23.52, 0.04]]]

Answer 2

你可以用collections.defaultdict来表示一个O（n）解。你知道吗

在Python3.7中，您将获得额外的好处，即值的顺序将与输入中的顺序相匹配。这在python3.6中可以工作，但被认为是一个实现细节。你知道吗

from collections import defaultdict

d = defaultdict(list)

for item in tideData:
    d[item[0]].append(item)

res = list(d.values())

结果：

[[['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73]],
 [['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]],
 [['Saturday 6 January', 0.5, 0.02], ['Saturday 6 January', 7.18, 0.85]],
 [['Friday 2 February', 23.52, 0.04]]]

对于那些对O（n）和O（n logn）解决方案之间的性能差异感兴趣的人：

from collections import defaultdict
from itertools import groupby

tideData = tideData * 10000

def jp(tideData):
    d = defaultdict(list)
    for item in tideData:
        d[item[0]].append(item)
    return list(d.values())

def grp(tideData):    
    return [list(g) for _,g in groupby(tideData, key=lambda x: x[0])]

%timeit jp(tideData)   # 12.2 ms per loop
%timeit grp(tideData)  # 33.1 ms per loop

Answer 3

这里有一个简单的方法，没有任何导入：

groub_by={}
for i,j in enumerate(tideData):
    if j[0] not in groub_by:
        groub_by[j[0]]=[j]
    else:
        groub_by[j[0]].append(j)
print(groub_by.values())

输出：

[[['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73]], [['Saturday 6 January', 0.5, 0.02], ['Saturday 6 January', 7.18, 0.85]], [['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]], [['Friday 2 February', 23.52, 0.04]]]