防止 OrderedDict 在迭代期间将单个字符串值拆分为字符

Question

我想创建一个OrderedDict，每个键对应一个字符串，每个值对应一个字符串列表。

from collections import OrderedDict

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

我需要通过访问键和值来遍历这个字典。在普通字典上，这很简单：

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

（如果这可以在理解中完成，请赐教——我无法弄清楚如何在不失去对密钥的访问权的情况下做到这一点。）

当我在OrderedDict 上尝试这个时，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

我做错了什么？

Answer 1

这是因为您将字符串 (("baz")) 分配给 b 键，而不是元组 (("baz",)) 或字符串列表。

代替

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

试试

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

（注意新的逗号。）

Answer 2

您可以在"baz" 之后添加,，以便在for 循环期间将其解释为元组而不是单个str：

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

然后，迭代可以很好地抓取元组 baz 的单个元素：

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

同样可以通过包装[]中的值来实现：

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

---

更好，如果您已经有了可用的字典，只需使用 ** 语法在 someOrderedDict 中展开（解压缩）字典 someDict：

someOrderedDict = OrderedDict(**someDict)

效果同样好，看起来更加紧凑和干净。