python中的子列表元素的計數和追加

Question

我想要計算子列表元素的唯一實例的數量，然後將每個唯一元素寫入新列表，並將子實例附加到子列表中。 list_1中的每個子列表都只有兩個元素，順序無關緊要。

這樣：

list_1 = [[a, b], [a, c], [a, c], [a, c], [b, e], [d, q], [d, q]] 

變爲：

new_list = [[a, b, 1], [a, c, 3], [b, e, 1], [d, q, 2]] 

我在想，我需要使用集，但我很欣賞的人指着我在正確的方向。

Answer 1

如果相同的子列表是連續的：

from itertools import groupby 

new_list = [sublist + [sum(1 for _ in g)] for sublist, g in groupby(list_1)] 
# -> [['a', 'b', 1], ['a', 'c', 3], ['b', 'e', 1], ['d', 'q', 2]] 

Answer 2

你想看看collections.Counter(); Counter物件爲多套（亦稱袋子）;他們將鍵映射到他們的計數。

您將得把你的子表到元組是可作爲鑰匙，但：

from collections import Counter 

counts = Counter(tuple(e) for e in list_1) 

new_list = [list(e) + [count] for e, count in counts.most_common()] 

，讓你通過計數排序的new_list（降序）：

>>> from collections import Counter 
>>> list_1 = [['a', 'b'], ['a', 'c'], ['a', 'c'], ['a', 'c'], ['b', 'e'], ['d', 'q'], ['d', 'q']] 
>>> counts = Counter(tuple(e) for e in list_1) 
>>> [list(e) + [count] for e, count in counts.most_common()] 
[['a', 'c', 3], ['d', 'q', 2], ['a', 'b', 1], ['b', 'e', 1]] 

如果您發生總是連續的，那麼你也可以使用itertools.groupby()：

from itertools import groupby 

def counted_groups(it): 
    for entry, group in groupby(it, key=lambda x: x): 
     yield entry + [sum(1 for _ in group)] 

new_list = [entry for entry in counted_groups(list_1)] 

我在這裏使用了一個單獨的生成器函數，但是可以將循環內聯到列表理解中。

這給：

>>> from itertools import groupby 
>>> def counted_groups(it): 
...  for entry, group in groupby(it, key=lambda x: x): 
...   yield entry + [sum(1 for _ in group)] 
... 
>>> [entry for entry in counted_groups(list_1)] 
[['a', 'b', 1], ['a', 'c', 3], ['b', 'e', 1], ['d', 'q', 2]] 

，並保留原始排序。

Answer 3

一個比特的 '輪的房屋' 溶液

list_1 = [['a', 'b'], ['a', 'c'], ['a', 'c'], ['a', 'c'], ['b', 'e'], ['d', 'q'], ['d', 'q']] 
new_dict={} 
new_list=[] 
for l in list_1: 
    if tuple(l) in new_dict: 
     new_dict[tuple(l)] += 1 
    else: 
     new_dict[tuple(l)] = 1 
for key in new_dict: 
    entry = list(key) 
    entry.append(new_dict[key]) 
    new_list.append(entry) 
print new_list