英文:
How do xor a dataframe slice with the next num and insert the answer in column 'dr'
问题
I will only translate the provided content without addressing your request. Here is the translation:
好的，我有这个数据帧，你会注意到它的名称是solve，我正在使用4的倍数切片。
在“rst”列中，我想要对1进行异或3并获得2（1^3=2）。然后我想要做3^7=4，我想要将这些值放在“dr”中的相应位置。所以，solve.loc[0:, ('dr')] = 2，以及solve.loc[4:, ('dr')] = 4。我的当前方法很繁琐，不够自动化，这是我正在做的：
In [13150]: np.array(solve.loc[::4, ('rst')]) ^ np.array(solve.loc[4::4, ('rst')])
ValueError: operands could not be broadcast together with shapes (16,) (15,)
这个错误可以通过以下解决：
In [13159]: wutwut = np.array(solve.loc[::4, ('rst')])[:15] ^ np.array(solve.loc[4::4, ('rst')])
Out[13159]:
array([ 2, 4, 2, 11, 2, 8, 0, 0, 7, 0, 6, 0, 8, 14, 2],
dtype=int8)
然后将这些值放回solve.loc['dr']是个问题，因为我必须手动设置长度，就像这样：
solve.loc[:56:4, ('dr')] = wutwut
如你所见，我必须手动设置长度，是否有更自动化的方式？
正如你所看到的，这很繁琐，不够实际，因为我正在处理不同和不断变化的长度，我需要一个更自动化的最佳实践方法。我正在寻求一些建议，提前感谢。此外，我还有更高级的用例，其中需要在列之间进行异或，如果有人有相关策略，对我以后的工作也会有帮助。
英文:
Ok I have this data frame which you notice is names solve and I'm using a slice of 4
In [13147]: solve[::4]
Out[13147]:
rst dr
0 1 0
4 3 0
8 7 0
12 5 0
16 14 0
20 12 0
24 4 0
28 4 0
32 4 0
36 3 0
40 3 0
44 5 0
48 5 0
52 13 0
56 3 0
60 1 0
What I want is to in column 'rst' xor 1 by 3 and get 2 (1^3=2). then I want to do 3^7 = 4, I want to put those in the corresponding spots in dr. so solve.loc[0:, ('dr')] = 2, and solve.loc[4:, ('dr')] = 4. My current method is tedious and not automatic, here is what I'm doing:
In [13150]: np.array(solve.loc[::4, ('rst')]) ^ np.array(solve.loc[4::4, ('rst')])
ValueError: operands could not be broadcast together with shapes (16,) (15,)
which is resolved with:
In [13159]: wutwut = np.array(solve.loc[::4, ('rst')])[:15] ^ np.array(solve.loc[4::4, ('rst')])
Out[13159]:
array([ 2, 4, 2, 11, 2, 8, 0, 0, 7, 0, 6, 0, 8, 14, 2],
dtype=int8)
and then putting the values back into solve.loc['dr'] is an issue because I have to bust a length in manually like:
solve.loc[:56:4, ('dr')] = wutwut
see I have to manually set the length, is there a more automatic way
As you can see this is tedious and not practical because I'm working with different and changing lengths and I need a more automatic best practice cast for this. I'm looking for some suggestions and thanks in advance. Also I have more advanced use cases where is xor between columns so if anyone has strategies for that it will help me down the road as well
答案1
得分: 2
Here is the translation of the provided code:
让我们使用 "loc" 进行索引和自动赋值：
s = solve.loc[::4, 'rst']
solve.loc展开收缩, 'dr'] = s[:-1].values ^ s[1:].values
rst dr
0 1 2
4 3 4
8 7 2
12 5 11
16 14 2
20 12 8
24 4 0
28 4 0
32 4 7
36 3 0
40 3 6
44 5 0
48 5 8
52 13 14
56 3 2
60 1 0
I hope this helps!
英文:
Lets use loc for indexing and automatic value assignment
s = solve.loc[::4, 'rst']
solve.loc展开收缩, 'dr'] = s[:-1].values ^ s[1:].values
rst dr
0 1 2
4 3 4
8 7 2
12 5 11
16 14 2
20 12 8
24 4 0
28 4 0
32 4 7
36 3 0
40 3 6
44 5 0
48 5 8
52 13 14
56 3 2
60 1 0
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