防止OrderedDict在迭代過程中將單個字符串值拆分為字符

Question

我想為每個鍵創建一個OrderedDict ，為每個值創建一個字符串列表。

from collections import OrderedDict

# Compare a dict with an OrderedDict
someDict = dict(a = ["foo","bar"], b = ["baz"])
someOrderedDict = OrderedDict([("a",("foo","bar")),("b",("baz"))])

我需要遍歷此字典以訪問鍵和值。 在普通字典中，這很簡單：

for k,v in someDict.items():
    for eachv in v:
        print("do something with key = " + k + " and value = " + eachv)
# returns the desired output:
# do something with key = a and value = foo
# do something with key = a and value = bar
# do something with key = b and value = baz

（如果可以理解的話，請告訴我-我不知道該怎么做而不失去對密鑰的訪問權。）

當我在OrderedDict上嘗試此操作時，我得到：

for k,v in someOrderedDict.items():
    for eachv in v:
        print("do something with key = " + k + " and value = " + eachv) 
# results in undesirable splitting of value "baz" into ["b","a","z"]:
# do something with key = a and value = foo
# do something with key = a and value = bar
# do something with key = b and value = b
# do something with key = b and value = a
# do something with key = b and value = z

我究竟做錯了什么？

Answer 1

這是因為您要為b鍵分配一個字符串（ ("baz") ）而不是一個元組（ ("baz",) ）或字符串列表。

代替

someOrderedDict = OrderedDict([("a",("foo","bar")),("b",("baz"))])

嘗試

someOrderedDict = OrderedDict([("a",("foo","bar")),("b",("baz",))])

（請注意新的逗號。）

Answer 2

您可以在"baz"之后添加一個,以便在for循環中將其解釋為元組而不是單個str ：

someOrderedDict = OrderedDict([("a",("foo","bar")),("b",("baz",))])

然后，迭代可以很好地獲取元組baz的單個元素：

for k,v in someOrderedDict.items():
    for eachv in v:
        print("do something with key = " + k + " and value = " + eachv)
do something with key = b and value = baz
do something with key = a and value = foo
do something with key = a and value = bar

通過將值包裝在[]可以實現相同的目的：

[("a", ["foo","bar"]),("b", ["baz"])]

---

更妙的是 ，如果你已經有了可用的字典，只是擴展（解壓縮）字典someDict在someOrderedDict利用**語法：

someOrderedDict = OrderedDict(**someDict)

它同樣運作良好， 看起來更緊湊和干凈 。