比较2字典最干净的方法是什么？ [英] What's the cleanest way to compare 2 dictionary?
问题描述
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,
推荐答案
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
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