字典中的差异

Question

我想弄明白一个字典（dict）里的差别，想知道是添加了什么，还是移除了什么，以及是从哪里来的。

这里有一个值被添加的例子：

original = {0: None, 1: False, 2: [16]}
new = {0: None, 1: False, 2: [2, 16]}

difference = True, {2: 2} # True = Added

这里有一个值被移除的例子：

original = {0: None, 1: False, 2: [16, 64]}
new = {0: None, 1: False, 2: [64]}

difference = False, {2: 16} # False = Removed

问题是我不知道怎么去获取这些差别。有没有人知道怎么才能做到这一点？

额外信息（不知道你们需不需要这些）：

• 这也可以适用于原始数据和新数据都是0或1的情况。
• 1和2不能同时激活。如果一个有值，另一个就是False。

Answer 1

这里有一个链接，里面有一个可以比较两个字典（也就是一组键值对）差异的函数，后面还有一些额外的评论和代码示例：

http://code.activestate.com/recipes/576644-diff-two-dictionaries/

下面是代码：

KEYNOTFOUND = '<KEYNOTFOUND>'       # KeyNotFound for dictDiff

def dict_diff(first, second):
    """ Return a dict of keys that differ with another config object.  If a value is
        not found in one fo the configs, it will be represented by KEYNOTFOUND.
        @param first:   Fist dictionary to diff.
        @param second:  Second dicationary to diff.
        @return diff:   Dict of Key => (first.val, second.val)
    """
    diff = {}
    # Check all keys in first dict
    for key in first.keys():
        if (not second.has_key(key)):
            diff[key] = (first[key], KEYNOTFOUND)
        elif (first[key] != second[key]):
            diff[key] = (first[key], second[key])
    # Check all keys in second dict to find missing
    for key in second.keys():
        if (not first.has_key(key)):
            diff[key] = (KEYNOTFOUND, second[key])
    return diff

Answer 2

我觉得这个代码挺容易理解的：

def dict_diff(left, right):
    diff = dict()
    diff['left_only'] = set(left) - set(right)
    diff['right_only'] = set(right) - set(left)
    diff['different'] = {k for k in set(left) & set(right) if left[k]!=right[k]}
    return diff

>>> d1 = dict(a=1, b=20, c=30, e=50)
>>> d2 = dict(a=1, b=200, d=400, e=500)
>>> dict_diff(d1, d2)
{'different': {'b', 'e'}, 'left_only': {'c'}, 'right_only': {'d'}}

Answer 3

正如我在其他问题中提到的，PyPI上有一个专门为这个任务准备的库，叫做datadiff库。这个库使用起来很简单，你可以利用它的输出结果来完成你需要做的事情。