Question

假设我们有一个词典listD列表，其中每个词典都嵌入了更多的词典。例如，假设listD的第一个元素是：

listD[0] = {"bar1":{"bar2":{"bar3":1234}}}

现在我想检查所有i的listD [i] [＆＃34; bar1＆＃34;] [＆＃34; bar2＆＃34;] [＆＃34; bar3＆＃34;] == 1234。对于i = 0的第一个元素，这很容易，因为我们可以使用表达式：

listD[0]["bar1"]["bar2"]["bar3"] == 1234

但我不能简单地编写一个循环：

for dictelem in listD:
  if dictelem["bar1"]["bar2"]["bar3"] == 1234:
    print "equals 1234"

这是因为listD的某些字典元素可能是

形式

listD[i] = {"bar1":{"bar2":"abcd"}} or
listD[i] = {"bar1":{"bar2":None}}

如果我尝试访问＆＃34; bar3＆＃34;当它不存在时，将引发错误。

现在我在代码中手动指定检查是否存在bar1，bar2和bar3键，以及它们是否实际上是字典。但这真的很冗长，而且我很确定这是一种更简单的方法，但我无法弄清楚如何做到这一点。

Answer 1

def dictcheck(d, p, v):
    if len(p):
        if isinstance(d,dict) and p[0] in d:
            return dictcheck(d[p[0]], p[1:], v)
    else:
        return d == v

您传递了一个字典d，一个键p路径，以及检查v的最终值。它将以递归方式进入dicts，最后检查最后一个值是否等于v。

>>> dictcheck({"bar1":{"bar2":{"bar3":1234}}}, ('bar1','bar2','bar3'), 1234)
True

>>> dictcheck({"bar1":1234}, ('bar1','bar2','bar3'), 1234)
False

所以，回答你的问题（我想检查listD [i] [“bar1”] [“bar2”] [“bar3”] == 1234 for all i ）： / p>

all(dictcheck(x, ('bar1','bar2','bar3'), 1234) for x in listD)

Answer 2

以这种方式使用try/except块：

for dictelem in listD:
    try:
        if dictelem["bar1"]["bar2"]["bar3"] == 1234:
            print "equals 1234"
    except TypeError:
        pass

Answer 3

When dealing with nested dictionaries, I think of them as a tree where the keys make up the path to the value. With that in mind, I created a non-recursive function, dict_path with takes in a nested dictionary, the key path, and a value in case not found:

def dict_path(dic, path, value_if_not_found=None):
    path = path.split('.')
    try:
        for key in path:
            dic = dic[key]
        return dic
    except (KeyError, TypeError):
        return value_if_not_found

listD = [
    {"bar1": {"bar2": 'abcd'}},
    {"bar1": {"bar2": None}},
    {"bar1": {"bar2": {"bar3": 1234}}},
]

for dic in listD:
    value = dict_path(dic, 'bar1.bar2.bar3')
    if value == 1234:
        print 'Equals 1234:', dic

The function keeps traversing the nested dictionary until one of these three conditions occur:

It found the value in question. In this case, return that value
Along the way, the object is a dictionary, but does not have the requested key. In this case, a KeyError is raised
The object is not a dictionary, the TypeError is raised

For case 2 and 3, we simply return value_if_not_found

Answer 4

你可以试试这个

for item in listD:
    if item.get("bar1",{}).get("bar2",{}).get("bar3","") == 1234:
        print "yeah, gotcha"

如何在Python中检查嵌套字典中的值？

4 个答案: