如何檢查第三個列表中是否存在兩個列表的元素組合？

Question

我有一個看起來像這樣的元組列表

data = [("hyd", "ab", 10.99), ("del", "ab", 12.99), ("del", "cc", 10.19), ("cal", "dd", 4.99), ("hyd", "ee", 13.11), ("noi", "dd", 10.49), ("noi", "bb", 10.99),]

我在下面還有另外兩個列表

loc = ["noi", "del", "cal", "hyd"]

dealer = ["ee", "dd", "ab", "cc", "bb"]

現在，對於dealer每個元素，我想要一個loc每個對應元素的值列表

因此，由於dealer有5個元素，因此我將為loc每個對應元素提供5個帶有值的列表。

就像是

對於ee ，它將再次檢查loc列表的每個元素，並從數據中找出loc每個元素包含的值

對於ee [None, None, None, 13.11]

因此我們可以看到ee上面檢查了noi ，在分配給None的values什么都沒有找到，然后它檢查了del ，沒有找到分配了None的東西，然后它對cal進行了檢查，發現了什么，分配了None，但對於hyd卻找到了13.11並因此分配了價值。

同樣的，
對於dd [10.49, None, 4.99, None] ，依此類推...

如何獲得針對經銷商五個要素的五個清單？

我試圖做這樣的事情

temp_list = []
for i in dealer:
   print("i", i)
   for j in loc:
      print("j", j)
      for k in data:
         #print("k", k)
         if i in k and j in k:
            temp_list.append(k[2])
         else:
            temp_list.append(None)

但是我沒有得到預期的輸出。 我如何獲得清單？

完成預期的輸出

ee [None, None, None, 13.11]
dd [10.49, None, 4.99, None]
ab [None, 12.99, None, 10.99]
cc [None, 10.99, None, None]
bb [10.99, None, None, None]

Answer 1

您可以以更高效的方式進行操作。 您的解決方案是O（len（data）* len（dealers）* len（locations））。 我們可以在O（len（data））中通過僅對數據進行一次迭代來完成：

data = [("hyd", "ab", 10.99), ("del", "ab", 12.99), ("del", "cc", 10.19), ("cal", "dd", 4.99), ("hyd", "ee", 13.11), ("noi", "dd", 10.49), ("noi", "bb", 10.99),]
locations = ["noi", "del", "cal", "hyd"]
dealers = ["ee", "dd", "ab", "cc", "bb"]


out = {dealer: [None] * len(loc) for dealer in dealers} 
loc_index = {val: index for index, val in enumerate(locations)}

for location, dealer, amount in data:
    try:
        out[dealer][loc_index[location]] = amount
    except (IndexError, KeyError):
        pass

print(out)
# {'ee': [None, None, None, 13.11], 'cc': [None, 10.19, None, None], 
# 'dd': [10.49, None, 4.99, None], 'ab': [None, 12.99, None, 10.99], 
# 'bb': [10.99, None, None, None]}

Answer 2

使用更好的數據結構！

假設data兩個元素在前兩個元素中不能相等，那么可以使用以下字典來簡化生活：

>>> from collections import defaultdict
>>> 
>>> data = [("hyd", "ab", 10.99), ("del", "ab", 12.99), ("del", "cc", 10.19), ("cal", "dd", 4.99), ("hyd", "ee", 13.11), ("noi", "dd", 10.49), ("noi", "bb", 10.99),]
>>> d = defaultdict(dict)
>>> 
>>> for key, subkey, val in data:
...:    d[key][subkey] = val
...:    
>>> d
>>> 
defaultdict(dict,
            {'cal': {'dd': 4.99},
             'del': {'ab': 12.99, 'cc': 10.19},
             'hyd': {'ab': 10.99, 'ee': 13.11},
             'noi': {'bb': 10.99, 'dd': 10.49}})

...因為現在您可以執行以下操作：

>>> loc = ["noi", "del", "cal", "hyd"]
>>> dealer = ["ee", "dd", "ab", "cc", "bb"]
>>> 
>>> [[d[lo].get(deal) for lo in loc] for deal in dealer]
>>> 
[[None, None, None, 13.11],
 [10.49, None, 4.99, None],
 [None, 12.99, None, 10.99],
 [None, 10.19, None, None],
 [10.99, None, None, None]]

...或者如果您想要一個dict ：

>>> {deal:[d[lo].get(deal) for lo in loc] for deal in dealer}
>>> 
{'ab': [None, 12.99, None, 10.99],
 'bb': [10.99, None, None, None],
 'cc': [None, 10.19, None, None],
 'dd': [10.49, None, 4.99, None],
 'ee': [None, None, None, 13.11]}