python - 如何使用 Python 比较来自同一字典键的嵌套字典值

Question

我有两个字典 Content_11 和 Content_05，在字典里面我有我需要比较的每个文件的校验和，如果校验和匹配打印类似成功的东西否则失败对于那个文件名。以下是我的数据结构，下面是我的代码片段。

Content_11 = {
    "controls": {
        "windows-library-1234.zip": "A123455adfasfasdfasdf",  # SHA 256 checksum
        "unix-library-1234.zip": "a2343dfasdfasdfasdfasdfasdfasdf"
    },
    "policies": {
        "oracle-1234.zip": "A123455adfasfasdfasdfad",
        "rhel7-1234.zip": "sdaf23234234234asdf",
    }
}

Content_05 = {
    "controls": {
        "windows-library-1234.zip": "A123455adfasfasdfasdf",
        "unix-library-1234.zip": "a2343dfasdfasdfasdfasdfasdfasdf"
    },
    "policies": {
        "oracle-1234.zip": "A123455adfasfasdfasdfad",
        "rhel7-1234.zip": "sdaf23234234234asdf",
    }
}

我查看了一些来自 stackoverflow 的问题，但没有找到与我相关的问题。任何建议或改进表示赞赏。

for key_05, value_05 in Content_05.items():  # iterating inside content_05 dict
    for key_05_1, value_05_1 in value_05.items():  # iterating inside the content_05 value for getting nested dict
        for key_011, value_011 in Content_11.items():  # iterating insde content_11 dict for comparison
            for key_11_1, value_11_1 in value_011.items():
                if key_05 == key_011:
                    if value_05_1 == value_11_1:
                        print "Key {} and its value is {} is matching with {} and hence Success".format(key_05_1,
                                                                                                        value_05_1,
                                                                                                        value_11_1)
                    else:
                        print "Key {} and its value is {} is not matching with {} and hence FAILURE".format(key_05_1,
                                                                                                        value_05_1,
                                                                                                        value_11_1)

Answer 1

你做的工作太多了；无需遍历两个词典，因为您可以只测试其中一个词典中的键是否在另一个词典中可用。

您可以使用 dict.get() 返回默认值并进一步简化测试；默认情况下 dict.get() 为缺失值返回 None，只需在比较中使用它:

for type_, checksums in Content_05.items():
    # type_ is assumed to be present in the other dictionary
    other_checksums = Content_11[type_]

    for filename, hash in checksums.items():
        other_hash = other_checksums.get(filename)
        if hash != other_hash:
            print("Failure: Checksum mismatch for {}, expected {}, got {}".format(
                filename, hash, other_hash))
        else:
            print("Success: {} checksum matched".format(filename))

我也尝试使用更易读的变量名；文件名、校验和和哈希比 key_05_1 等更容易理解。