检查相应列表中是否存在值

Question

我正在尝试创建一个 for 循环，在其中动态检查相应列表中是否存在某些值。我不知道是否可以转换列表中的字符串，或者是否有更好的方法来做到这一点。

rating_1 = ['no', 'yes']
rating_2 = ['no', 'yes']

for item in d:
    if d[item] not in item: # I don't want to use the item,
                            # only get name that will match the respective list above
        print "value not allowed"

---

d =  {'rating_2': u'no', 'rating_1': u'no'}

Answer 1

✓ @anvd，我从您的问题中了解到，您想在列表 rating1 中搜索字典 d 的值是否存在& 评级2。

如果我错了或者我在下面提供的解决方案不能满足您的需求，请发表评论。

我建议您再创建 1 个字典 d_lists，将列表名称映射到原始列表对象。

步骤：

✓ 从 d 中获取每个键（列表名称）。

✓ 在 d_lists 中查找此键的存在。

✓ 从 d_lists 中获取对应的列表对象。

✓ 在选取的列表对象中找到 d 中键指向的值的存在。

✓ 如果找到元素，则停止迭代并搜索 d 中是否存在下一个值。

✓ 打印相关消息。

这是您修改后的代码（稍作修改）

我在下面的代码示例之后展示了另一个很好的示例。

rating_1 = ['nlo', 'yes']
rating_2 = ['no', 'yes']

# Creating a dictionary that maps list names to themselves (original list object)
d_lists = {
    "rating_1": rating_1,
    "rating_2": rating_2,
}

# Creating a dictionary that maps list to the item to be searched
# We just want to check whether 'rating_1' & 'rating_2' contains 'no' or not
d =  {'rating_2': u'no', 'rating_1': u'no'}

# Search operation using loop
for list_name in d:
    if list_name in d_lists:
        found = False
        for item in d_lists[list_name]:
            if item == d[list_name]:
                found = True
                break
        if found:
            print "'" + d[list_name] + "' exists in", d_lists[list_name];
        else:
            print "'" + d[list_name] + "' doesn't exist in", d_lists[list_name]
    else:
        print "Couldn't find", list_name

输出：

'no' exists in ['no', 'yes']
'no' doesn't exist in ['nlo', 'yes']

现在，看看另一个例子。

另一个大例子：

rating_1 = ['no', 'yes', 'good', 'best']
rating_2 = ['no', 'yes', 'better', 'worst', 'bad']
fruits = ["apple", "mango", "pineapple"]

# Creating a dictionary that maps list names to themselves (original list object)
d_lists = {
    "rating_1": rating_1,
    "rating_2": rating_2,
    "fruits": fruits,
}

# Creating a dictionary that maps list to the item to be searched
# We just want to check whether 'rating_1' & 'rating_2' contains 'no' or not
d =  {'rating_2': u'best', 'rating_1': u'good', 'fruits2': "blackberry"}

# Search operation using loop
for list_name in d:
    if list_name in d_lists:
        print "Found list referred by key/name", list_name, "=", d_lists[list_name];
        found = False
        for item in d_lists[list_name]:
            if d[list_name] == item:
                found = True
                break
        if found:
            print "'" + d[list_name] + "' exists in", d_lists[list_name], "\n";
        else:
            print "'" + d[list_name] + "' doesn't exist in", d_lists[list_name], "\n"
    else:
        print "Couldn't find list referred by key/name ", list_name, "\n"

输出：

Found list referred by key/name rating_2 = ['no', 'yes', 'better', 'worst', 'bad']
'best' doesn't exist in ['no', 'yes', 'better', 'worst', 'bad'] 

Couldn't find list referred by key/name  fruits2 

Found list referred by key/name rating_1 = ['no', 'yes', 'good', 'best']
'good' exists in ['no', 'yes', 'good', 'best'] 

Answer 2

您可以对允许的值列表使用另一个映射：

d =  {'rating_2': 'no', 'rating_1': 'no'}
allowed_values = {'rating_2': ['no', 'yes'], 'rating_1': ['no', 'yes']}

is_valid = all(d[item] in allowed_values[item] for item in d)

invalid_items = {k: v for k, v in d.items() if v not in allowed_values[k]}

Answer 3

my_lists = {
'rating_1' = ['no', 'yes'],
'rating_2' = ['no', 'yes'],
}

d =  {'rating_2': u'no', 'rating_1': u'no'}

for item in d:
    if d[item] not in my_list[item]:
        print "value not allowed"

或者，如果您想使用变量，请使用vars()，它提供了当前命名空间的字典，您可以在其中使用变量名称作为键。

rating_1 = ['no', 'yes']
rating_2 = ['no', 'yes']

d =  {'rating_2': u'no', 'rating_1': u'no'}

for item in d:
    if d[item] not in vars()[item]:
        print "value not allowed"

Answer 4

您应该为可变数量的变量使用字典。假设您正在寻找执行某种验证，您可以创建一个无效项目的字典。一种方法是通过迭代视图dict.items：

d =  {'rating_2': 'noo', 'rating_1': 'no'}
allowed_values = {'rating_2': ['no', 'yes'], 'rating_1': ['no', 'yes']}

bad_items = {}

for k, v in d.items():  
    if v not in allowed_values[k]:
        bad_items[k] = v

print(bad_items)

{'rating_2': 'noo'}

另一种 Pythonic 方法是使用字典推导：

bad_items = {k: v for k, v in d.items() if v not in allowed_values[k]}