添加串列作為字典值

如果我有一個串列串列，例如

l=[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]

和一個字典，其值本身就是字典型別，例如

d={"id1":{"A": "", "C": "", "G": "", "T": ""},"id2":{"A":"","C":"","G":"","T":""},
"id3":{"A": "", "C": "", "G": "", "T": ""},"id4":{"A": "", "C": "", "G": "", "T": ""}}

我想遍歷串列和每個專案到 d 中的相應鍵，輸出就像

d={"id1":{"A": "1", "C": "2", "G": "3", "T": "4"},"id2":{"A":"5","C":"6","G":"7","T":"8"},
"id3":{"A": "9", "C": "10", "G": "11", "T": "12"},"id4":{"A": "13", "C": "14", "G": "15", "T": "16"}}

但相反，它給了我以下

{'id1': {'A': 15, 'C': 15, 'G': 15, 'T': 15}, 'id2': {'A': 15, 'C': 15, 'G': 15, 'T': 15}, 'id3': {'A': 15, 'C': 15, 'G': 15, 'T': 15}, 'id4': {'A': 15, 'C': 15, 'G': 15, 'T': 15}}

到目前為止我的代碼

for k,v in d.items():
    for sub_k,sub_value in v.items():
        for i in range(len(l)):
            for j in range(len(l[i])):
                d[k][sub_k]=l[i][j]

uj5u.com熱心網友回復：

您的代碼的問題是您在每個級別的 dict 上迭代兩次，因此，您看到的結果是最后一次迭代的結果。您可以通過執行以下操作來修復您的代碼：

for i, (k,v) in enumerate(d.items()):
    for j, (sub_k,sub_value) in enumerate(v.items()):
         d[k][sub_k]=str(l[i][j]) # adding str() allows to change the int to a str

enumerate允許您迭代您的 dict（例如 k，v）并訪問索引（例如 i）

uj5u.com熱心網友回復：

這是我的想法：

new_dict = list(zip(d.keys(), d.values(), l))
updated_dic = {}
for i in new_dict:
    updated_dic[i[0]] = (dict(zip(i[1], i[2])))

輸出：

{'id1': {'A': 1, 'C': 2, 'G': 3, 'T': 4}, 'id2': {'A': 5, 'C': 6, 'G': 7, 'T': 8}, 'id3': {'A': 9, 'C': 10, 'G': 11, 'T': 12}, 'id4': {'A': 13, 'C': 14, 'G': 15, 'T': 16}}

uj5u.com熱心網友回復：

由于長度相匹配的，而是采用了雙for回圈，可以zip d與l到您遍歷一次創建的元組。使用dict壓縮迭代（內部字典鍵和子串列）上的建構式來構造內部字典。

for (Id, dct), vals in zip(d.items(), l):
    d[Id] = dict(zip(dct, vals))

這可以更簡潔地寫成字典理解：

d = {Id: dict(zip(dct, vals)) for (Id, dct), vals in zip(d.items(), l)}

輸出：

{'id1': {'A': 1, 'C': 2, 'G': 3, 'T': 4},
 'id2': {'A': 5, 'C': 6, 'G': 7, 'T': 8},
 'id3': {'A': 9, 'C': 10, 'G': 11, 'T': 12},
 'id4': {'A': 13, 'C': 14, 'G': 15, 'T': 16}}

uj5u.com熱心網友回復：

這將為您完成

l=[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
d={"id1":{"A": "", "C": "", "G": "", "T": ""},"id2":{"A":"","C":"","G":"","T":""},
"id3":{"A": "", "C": "", "G": "", "T": ""},"id4":{"A": "", "C": "", "G": "", "T": ""}}

dl = [inner.keys() for inner in d.values()] # get a list of lists of the keys
print({f"id{i 1}": dict(zip(*p)) for i, p in enumerate(zip(dl, l))}) 
# first zip the sublists of values with the sublist of keys
# then zip the tuple of sublists of values and sublist of keys by unpacking it using * 
# we get something like (("a", 1), ("b", 2)) which can be converted to a dict
# finally use dict comprehension to puth the id's back

但它相當難讀，所以需要評論

uj5u.com熱心網友回復：

整個代碼應如下所示：

d = {"id1":{"A": "", "C": "", "G": "", "T": ""},"id2":{"A":"","C":"","G":"","T":""},
"id3":{"A": "", "C": "", "G": "", "T": ""},"id4":{"A": "", "C": "", "G": "", "T": ""}}
l = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]

for i, (k, v) in enumerate(d.items()):
    for j, (k2, value) in enumerate(v.items()):
         d[k][k2]=str(l[i][j])

print (d)

它解決了字串格式和放入字典中的值

uj5u.com熱心網友回復：

l = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
d = {"id1": {"A": "", "C": "", "G": "", "T": ""}, "id2": {"A": "", "C": "", "G": "", "T": ""},
     "id3": {"A": "", "C": "", "G": "", "T": ""}, "id4": {"A": "", "C": "", "G": "", "T": ""}}

for index, key in enumerate(d.keys()):
    for new_key, val in zip(d[key].keys(), l[index]):
        d[key][new_key] = str(val)

print(d)

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