元組= [['nike airjordan kahverengi', 42, 6],['nike airjordan sar?', 42, 10], ['nike airjordan mavi', 42, 8],['nike airjordan ye?il', 42, 9] ['nike airjordan sar?', 42, 10]]
def getOrderList(self):
orderSplit = ''.join(self.orderText).splitlines()
productList = []
for idx, i in enumerate(orderSplit, start = 0):
if(idx > self.starting_point):
if i == '@':
break
else:
productList.append([i.rstrip(i[-3:]), int(i[-2:]),idx])
return productList
這是獲取元組的函式
orderList = self.product.getOrderList()
在這里，我從現在開始在元組中尋找相同的
uj5u.com熱心網友回復：
您可以將串列映射到元組并使用collections.Counter：
lsts = [['nike airjordan kahverengi', 42, 6],['nike airjordan sar?', 42, 10],
['nike airjordan mavi', 42, 8],['nike airjordan ye?il', 42, 9],
['nike airjordan sar?', 42, 10]]
from collections import Counter
counts = Counter(map(tuple, lsts))
for k,v in counts.items():
print(list(k), 'appears', v, 'time(s).')
輸出：
['nike airjordan kahverengi', 42, 6] appears 1 time(s).
['nike airjordan sar?', 42, 10] appears 2 time(s).
['nike airjordan mavi', 42, 8] appears 1 time(s).
['nike airjordan ye?il', 42, 9] appears 1 time(s).
?
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