我有两个这样的数组：
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Content="{Binding}"
ContentTemplate="{TemplateBinding MainTemplate}"
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其中第一列包含标识符，其余列包含一些数据，其中B的列数远大于A的列数。标识符是唯一的。 A中的行数可以小于B中的行数，因此在某些情况下需要空的间隔行 我正在寻找一种有效的方法来匹配矩阵A的行到矩阵B，以便结果看起来像这样：
A = [[111, ...], B = [[222, ...],
[222, ...], [111, ...],
[333, ...], [333, ...],
[555, ...]] [444, ...],
[555, ...]]
我可以对两个矩阵进行排序或编写for循环，但这两种方法看起来都很笨拙......是否有更好的实现？
答案 0 :(得分：2)
这是使用np.searchsorted -
# Store the sorted indices of A
sidx = A[:,0].argsort()
# Find the indices of col-0 of B in col-0 of sorted A
l_idx = np.searchsorted(A[:,0],B[:,0],sorter = sidx)
# Create a mask corresponding to all those indices that indicates which indices
# corresponding to B's col-0 match up with A's col-0
valid_mask = l_idx != np.searchsorted(A[:,0],B[:,0],sorter = sidx,side='right')
# Initialize output array with NaNs.
# Use l_idx to set rows from A into output array. Use valid_mask to select
# indices from l_idx and output rows that are to be set.
out = np.full((B.shape[0],A.shape[1]),np.nan)
out[valid_mask] = A[sidx[l_idx[valid_mask]]]
请注意，valid_mask也可以使用np.in1d：np.in1d(B[:,0],A[:,0])创建，以获得更直观的答案。但是，我们正在使用np.searchsorted，因为它在性能方面更好，在this other solution中也有更详细的讨论。
示例运行 -
In [184]: A
Out[184]:
array([[45, 11, 86],
[18, 74, 59],
[30, 68, 13],
[55, 47, 78]])
In [185]: B
Out[185]:
array([[45, 11, 88],
[55, 83, 46],
[95, 87, 77],
[30, 9, 37],
[14, 97, 98],
[18, 48, 53]])
In [186]: out
Out[186]:
array([[ 45., 11., 86.],
[ 55., 47., 78.],
[ nan, nan, nan],
[ 30., 68., 13.],
[ nan, nan, nan],
[ 18., 74., 59.]])
答案 1 :(得分：0)
简单方法是从dict构建A，然后使用它将B中找到的标识符映射到新数组。
构建dict：
>>> A = [[1,"a"], [2,"b"], [3,"c"]]
>>> A_dict = {x[0]: x for x in A}
>>> A_dict
{1: [1, 'a'], 2: [2, 'b'], 3: [3, 'c']}
映射：
>>> B = [[3,"..."], [2,"..."], [1,"..."]]
>>> result = (A_dict[x[0]] for x in B)
>>> list(result)
[[3, 'c'], [2, 'b'], [1, 'a']]
答案 2 :(得分：0)
如果您希望将B中的值连接到A，则不清楚。让我们假设不...那么最简单的方法可能只是建立一个标识符的字典行，然后重新排序A：
def match_order(A, B):
# identifier -> row
by_id = {A[i, 0]: A[i] for i in range(len(A))}
# make up a fill row and rearrange according to B
fill_row = [-1] * A.shape[1]
return numpy.array([by_id.get(k, fill_row) for k in B[:, 0]])
例如，如果我们有：
A = numpy.array([[111, 1], [222, 2], [333, 3], [555, 5]])
B = numpy.array([[222, 2], [111, 1], [333, 3], [444, 4], [555, 5]])
然后
>>> match_order(A, B)
array([[222, 2],
[111, 1],
[333, 3],
[ -1, -1],
[555, 5]])
如果您希望连接B，那么您可以这样做：
>>> numpy.hstack( (match_order(A, B), B[:, 1:]) )
array([[222, 2, 2],
[111, 1, 1],
[333, 3, 3],
[ -1, -1, 4],
[555, 5, 5]])
答案 3 :(得分：0)
>>> A = [[3,'d', 'e', 'f'], [1,'a','b','c'], [2,'n','n','n']]
>>> B = [[1,'a','b','c'], [3,'d','e','f']]
>>> A_dict = {x[0]:x[1:] for x in A}
>>> A_dict
{1: ['a', 'b', 'c'], 2: ['n', 'n', 'n'], 3: ['d', 'e', 'f']}
>>> B_dict = {x[0]:x[1:] for x in B}
>>> B_dict
{1: ['a', 'b', 'c'], 3: ['d', 'e', 'f']}
>>> result=[[x] + A_dict[x] for x in A_dict if x in B_dict and A_dict[x]==B_dict[x]]
>>> result
[[1, 'a', 'b', 'c'], [3, 'd', 'e', 'f']]
这里A [0]，B [1]和A [1]，B [0]是相同的。转换成字典并处理问题可以使这里的生活变得更轻松。
步骤1：为每个2D列表创建dict对象。
步骤2：迭代A_dict中的每个键并检查： 一个。如果Key存在于B_dict中， 湾如果是，请查看两个键是否具有相同的值
步骤3：附加键和值以形成二维列表。
干杯！