比较2字典最干净的方法是什么？

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嗨列表， 我相信有很多方法可以进行比较但是我喜欢看看 如果你有2个字典你会怎么做套装（包含大量的 数据 - 比如20000个密钥，每个密钥包含十几个记录） 你想建立一个关于这两个集的差异列表。 我最终得到3个列表：A中有什么而B中没有，B中有什么而不是A，当然，A和B都是什么。 您认为最干净的方法是什么？ （我相信你会 想出令我惊讶的方式：=）） 谢谢，

Hi list, I am sure there are many ways of doing comparision but I like to see what you would do if you have 2 dictionary sets (containing lots of data - like 20000 keys and each key contains a dozen or so of records) and you want to build a list of differences about these two sets. I like to end up with 3 lists: what''s in A and not in B, what''s in B and not in A, and of course, what''s in both A and B. What do you think is the cleanest way to do it? (I am sure you will come up with ways that astonishes me :=) ) Thanks,

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John Henry写道： John Henry wrote: 嗨列表， 我相信有很多方法可以进行比较，但我喜欢看看 如果你有2个字典集（包含很多 $ b）你会怎么做$ b数据 - 比如20000个密钥，每个密钥包含十几个记录） 你想建立一个关于这两个集的差异列表。 我最终得到了3个列表：A中有什么而不​​是B中的内容，B中有什么内容，而不是A中的内容，当然，还有什么内容A和B都是。 您认为最干净的方法是什么？ （我相信你会 想出令我惊讶的方式：=）） 谢谢， Hi list, I am sure there are many ways of doing comparision but I like to see what you would do if you have 2 dictionary sets (containing lots of data - like 20000 keys and each key contains a dozen or so of records) and you want to build a list of differences about these two sets. I like to end up with 3 lists: what''s in A and not in B, what''s in B and not in A, and of course, what''s in both A and B. What do you think is the cleanest way to do it? (I am sure you will come up with ways that astonishes me :=) ) Thanks,

我赚4个箱子： a_exclusive_keys b_exclusive_keys common_keys_equal_values common_keys_diff_values 类似于： a = {1：1,2：2,3：3,4：4} b = {2：2,3：-3,5：5} keya = set（a.keys（）） keyb = set（b.keys（） ） a_xclusive = keya - keyb b_xclusive = keyb - keya _common = keya& keyb common_eq = set（如果a [k] == b [k]，则_com中k为k） common_neq = _common - common_eq 如果你现在简单设置算术，它应该是OK。 - Paddy。

I make it 4 bins: a_exclusive_keys b_exclusive_keys common_keys_equal_values common_keys_diff_values Something like: a={1:1, 2:2,3:3,4:4} b = {2:2, 3:-3, 5:5} keya=set(a.keys()) keyb=set(b.keys()) a_xclusive = keya - keyb b_xclusive = keyb - keya _common = keya & keyb common_eq = set(k for k in _common if a[k] == b[k]) common_neq = _common - common_eq If you now simple set arithmatic, it should read OK. - Paddy.

Paddy写道： Paddy wrote: John Henry写道： John Henry wrote: 嗨列表， 我相信有很多方法可以进行比较但是我喜欢看看 如果你有2个字典集（包含很多）你会怎么做数据 - 比如20000个密钥，每个密钥包含十几个记录） ，你想建立一个关于这两个集的差异列表。 我最终得到3个列表：A中有什么而B中没有，B中有什么而不是A，当然还有什么''在A和B中都有。 您认为最干净的方法是什么？ （我相信你会 想出令我惊讶的方式：=）） 谢谢， Hi list, I am sure there are many ways of doing comparision but I like to see what you would do if you have 2 dictionary sets (containing lots of data - like 20000 keys and each key contains a dozen or so of records) and you want to build a list of differences about these two sets. I like to end up with 3 lists: what''s in A and not in B, what''s in B and not in A, and of course, what''s in both A and B. What do you think is the cleanest way to do it? (I am sure you will come up with ways that astonishes me :=) ) Thanks,

我赚4个箱子： a_exclusive_keys b_exclusive_keys common_keys_equal_values common_keys_diff_values 类似于： a = {1：1,2：2,3：3,4：4} b = {2：2,3：-3,5：5} keya = set（a.keys（）） keyb = set（b.keys（） ） a_xclusive = keya - keyb b_xclusive = keyb - keya _common = keya& keyb common_eq = set（如果a [k] == b [k]，则_com中k为k） common_neq = _common - common_eq 如果你现在简单设置算术，它应该是OK。 - Paddy。

I make it 4 bins: a_exclusive_keys b_exclusive_keys common_keys_equal_values common_keys_diff_values Something like: a={1:1, 2:2,3:3,4:4} b = {2:2, 3:-3, 5:5} keya=set(a.keys()) keyb=set(b.keys()) a_xclusive = keya - keyb b_xclusive = keyb - keya _common = keya & keyb common_eq = set(k for k in _common if a[k] == b[k]) common_neq = _common - common_eq If you now simple set arithmatic, it should read OK. - Paddy.

谢谢，这很干净。给我一个很好的理由升级到Python 2.4。

Thanks, that''s very clean. Give me good reason to move up to Python 2.4.

John Henry写道： John Henry wrote: Paddy写道： Paddy wrote: John Henry写道： John Henry wrote: 嗨列表， > 我相信有很多方法可以做比较但是我喜欢看 是什么如果你有2个字典集（包含很多 数据 - 就像20000个密钥，每个密钥包含十几个记录）你会怎么做？ 你想建立关于这两组的差异列表。 > 我最终得到3个列表：A中有什么而B中没有，是什么''在B 而不在A中，当然，A和B都有。 > 做什么你认为这是最干净的方法吗？ （我相信你会 想出令我惊讶的方式：=）） > 谢谢， Hi list, > I am sure there are many ways of doing comparision but I like to see what you would do if you have 2 dictionary sets (containing lots of data - like 20000 keys and each key contains a dozen or so of records) and you want to build a list of differences about these two sets. > I like to end up with 3 lists: what''s in A and not in B, what''s in B and not in A, and of course, what''s in both A and B. > What do you think is the cleanest way to do it? (I am sure you will come up with ways that astonishes me :=) ) > Thanks,

我把它变成4个箱子： a_exclusive_keys b_exclusive_keys common_keys_equal_values common_keys_diff_values 类似于： a = {1：1,2：2,3：3,4：4} b = {2：2,3：-3,5：5} keya = set（a.keys（）） keyb = set （b.keys（）） a_xclusive = keya - keyb b_xclusive = keyb - keya _common = keya& keyb common_eq = set（如果a [k] == b [k]，则_com中k为k） common_neq = _common - common_eq 如果你现在简单设置算术，那就应该看好了。 - 帕迪。

I make it 4 bins: a_exclusive_keys b_exclusive_keys common_keys_equal_values common_keys_diff_values Something like: a={1:1, 2:2,3:3,4:4} b = {2:2, 3:-3, 5:5} keya=set(a.keys()) keyb=set(b.keys()) a_xclusive = keya - keyb b_xclusive = keyb - keya _common = keya & keyb common_eq = set(k for k in _common if a[k] == b[k]) common_neq = _common - common_eq If you now simple set arithmatic, it should read OK. - Paddy.

谢谢，这很干净。给我一个很好的理由升级到Python 2.4。

Thanks, that''s very clean. Give me good reason to move up to Python 2.4.

哦，等等，也适用于2.3。 只需要： from sets import设置

Oh, wait, works in 2.3 too. Just have to: from sets import Set as set