Question

我有一个元组列表：

[(0, 1), (0, 1), (0, 0), (0, 0), (1, 0), (1, 0), (1, 1), (1, 0), (1, 0), (2, 0), (2, 1), (2, 0), (3, 0), (3, 1), (3, 1), (3, 0), (3, 0), (4, 0), (4, 1), (4, 0), (4, 1), (4, 1), (5, 0), (5, 0), (5, 1), (5, 1)]

我想求和元组的右边等于左边的元组，并将其放在另一个元组列表中，因此对于上面的列表，我将得到：

[(0,2),(1,1),(2,1),(3,2),(4,3),(5,2)]

我尝试过：

k=0
for i,TCtup in enumerate(wordsMatchingList):
    if wordsMatchingList[i][0]==k:
        TC_matches+=wordsMatchingList[i][1]
        print("k: {} /// TC_matches: {}".format(k,TC_matches)) #for checking
    else:
        groupedWordsMatchingList.append(tuple((k,TC_matches)))
        TC_matches=0
        k+=1

但是从k = 1开始，由于else条件，它每k循环少一遍。

谢谢

Answer 1

如果保证您的元组按这样的顺序排列-全部(0, x)，然后所有(1, x)，等等-您可以使用groupby：

>>> xs = [(0, 1), (0, 1), (0, 0), (0, 0), (1, 0), (1, 0), (1, 1), (1, 0), (1, 0), (2, 0), (2, 1), (2, 0), (3, 0), (3, 1), (3, 1), (3, 0), (3, 0), (4, 0), (4, 1), (4, 0), (4, 1), (4, 1), (5, 0), (5, 0), (5, 1), (5, 1)]
>>> from itertools import groupby
>>> from operator import itemgetter
>>> groups = groupby(xs, key=itemgetter(0))
>>> ys = [(key, sum(map(itemgetter(1), group))) for key, group in groups]

如果它们不是，但是您可以对它们进行排序（您有一个列表，不仅是一个任意的可迭代对象，而且它不是很大，以至于对数线性时间会太昂贵）：

>>> groups = groupby(sorted(xs, key=itemgetter(0)), key=itemgetter(0))

如果无法对它们进行排序，则可以在进行过程中手动建立总计：

>>> from collections import Counter
>>> totals = Counter()
>>> for k, v in xs:
...     totals[k] += v
>>> ys = list(totals.items())

Answer 2

另一种方法是使用collections中的defaultdict并迭代元组列表。

from collections import defaultdict

lst = [(0, 1), (0, 1), (0, 0), (0, 0), (1, 0), (1, 0), (1, 1), (1, 0), (1, 0), (2, 0), (2, 1), (2, 0), (3, 0), (3, 1), (3, 1), (3, 0), (3, 0), (4, 0), (4, 1), (4, 0), (4, 1), (4, 1), (5, 0), (5, 0), (5, 1), (5, 1)]

d = defaultdict(int)

for (u,v) in lst:
    d[u]+=v

# list(d.items()) [(0, 2), (1, 1), (2, 1), (3, 2), (4, 3), (5, 2)]

Answer 3

另一种方式，

t.sort(key=lambda x: x[0]) #sort before groupby (required)
g=itertools.groupby(t, lambda x: x[0])
new_l = []
for k,v in g:
    new_l.append((k, sum([x[1] for x in v])))

Answer 4

我建议使用带有groupby函数的库。例如，pandas可能有用

>>> s = pd.DataFrame(list_)
>>> s.groupby(0, as_index=False).sum().agg(tuple,1).tolist()

[(0, 2), (1, 1), (2, 1), (3, 2), (4, 3), (5, 2)]

Answer 5

In [5]: [(j, sum([i[1] for i in a if i[0] == j])) for j in set([i[0] for i in a])]
Out[5]: [(0, 2), (1, 1), (2, 1), (3, 2), (4, 3), (5, 2)]

Answer 6

lst = [(0, 1), (0, 1), (0, 0), (0, 0), (1, 0), (1, 0), (1, 1), (1, 0), (1, 0), (2, 0), (2, 1), (2, 0), (3, 0), (3, 1), (3, 1), (3, 0), (3, 0), (4, 0), (4, 1), (4, 0), (4, 1), (4, 1), (5, 0), (5, 0), (5, 1), (5, 1)]
[(i,sum([q[1] for q in lst if q[0] == i])) for i in range(lst[-1][0]+1)]

给予：

[(0,2),(1,1),(2,1),(3,2),(4,3),(5,2)]

汇总列表中的元组

6 个答案: