[英]Python 3 list comprehension sum and concatenation
mylist =[[16, 'August', 2014, 540, 10], [16, 'Iunie', 2014, 100, 20], [23,'August', 2014, 540,10], [23,'Septembrie', 2016, 540,30], [21,'August', 2014, 422, 30]]
我想要實現的是:
manipulate = [[16, 'August 2014, Iunie 2014', 540, 30], [23, 'August 2014, Septembrie 2016', 540, 40], [21, 'August', 2014, 422, 30]]
並解釋一下,在mylist(列表列表)中,我想檢查索引0(第一個元素)的每個列表,當它找到相同的值元素時(在我的例子中16 === 16)並執行以下操作:keep只有第一個列表並累積到索引1 +索引2(如此示例 - >'8月',2014年和'Iunie',2014年到'2014年8月,Iunie 2014')並將索引4加起來(如我的示例中所示20) + 10)。
我設法這樣做:
manipulate=[i for i in mylist if i[0] not in [mylist[idx][0] for idx in range(0,mylist.index(i))]]
但這將輸出:
[[16, 'August', 2014, 540, 10], [23, 'August', 2014, 540, 10], [21, 'August', 2014, 422, 30]]
為了也做我想要的計算,我必須添加什么?,我使用python 3.非常感謝你!
假設列表已經按第一個元素排序,您可以使用itertools.groupby
對元素進行分組,然后組合新的子列表。
manipulate = []
for group in (list(g) for k, g in itertools.groupby(mylist, key=lambda x: x[0])):
lst = [group[0][0]] + [x for g in group for x in g[1:3]] + [group[0][3]] + [sum(g[-1] for g in group)]
manipulate.append(lst)
如果你想要的話,你甚至可以把它們放在一個可怕的列表中...
manipulate = [[group[0][0]] + [x for g in group for x in g[1:3]] + [group[0][3]] + [sum(g[-1] for g in group)]
for group in (list(g) for k, g in itertools.groupby(mylist, key=lambda x: x[0]))]
結果就是這個(我不合並字符串,因為我不確定你真的是這個意思):
[[16, 'August', 2014, 'Iunie', 2014, 540, 30],
[23, 'August', 2014, 'Septembrie', 2016, 540, 40],
[21, 'August', 2014, 422, 30]]
如果你想合並日期,請使用這個稍微更糟糕的列表理解:
manipulate = [[group[0][0]] + [', '.join(' '.join(map(str, g[1:3])) for g in group)] + [group[0][3]] + [sum(g[-1] for g in group)]
for group in (list(g) for k, g in itertools.groupby(mylist, key=lambda x: x[0]))]
這樣,結果是:
[[16, 'August 2014, Iunie 2014', 540, 30],
[23, 'August 2014, Septembrie 2016', 540, 40],
[21, 'August 2014', 422, 30]]
這適用於每組任意數量的元素。 如果子列表未按第一個元素排序,則首先對它們進行排序,或使用字典進行分組。
這是非常冗長但我覺得它必須是, 可理解\\可讀 :
mylist = [[16, 'August', 2014, 540, 10], [16, 'Iunie', 2014, 100, 20], [23, 'August', 2014, 540, 10], [23, 'Septembrie', 2016, 540, 30], [21, 'August', 2014, 422, 30]]
g = {}
for x, *y in mylist:
g.setdefault(x, []).append(y)
def formater(obj, a_list):
if len(a_list) == 1:
return [obj] + a_list[0]
else:
conc = ', '.join('{} {}'.format(sub[0], sub[1]) for sub in a_list)
return [obj, conc, a_list[0][2], sum(sub[3] for sub in a_list)]
manipulate = [formater(k, v) for k, v in g.items()]
print(manipulate) # [[16, 'August 2014, Iunie 2014', 540, 30], [21, 'August', 2014, 422, 30], [23, 'August 2014, Septembrie 2016', 540, 40]]
你可以試試這個:
mylist =[[16, 'August', 2014, 540, 10], [16, 'Iunie', 2014, 100, 20], [23,'August', 2014, 540,10], [23,'Septembrie', 2016, 540,30], [21,'August', 2014, 422, 30]]
new_list = [[mylist[i][0]]+[mylist[i][1]+" "+str(mylist[i][2])]+mylist[i][3:] for i in range(len(mylist))]
final_lists = [[c+", "+d if isinstance(c, str) and isinstance(d, str) else c+d if c != d else c for c, d in zip(new_list[i], new_list[i+1])] for i in range(0, len(new_list)-1, 2)]+mylist[-1]
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