[英]How to convert three lists x,y,z into a matrix z[x,y]?
Given three lists x,y,z
of identical size à la 给定三个相同大小的列表
x,y,z
x = [1, 0,.2,.2, 1, 0]
y = [0, 0, 0, 1,.2,.2]
z = [0, 2, 3, 1, 0, 1]
with unique but incomplete pairings of x,y
float values, how to map z
to a matrix Z[i,j]
where i,j
correspond to the indices np.unique
of x,y
respectively? 具有唯一但不完整的
x,y
浮点值对,如何将z
映射到矩阵Z[i,j]
,其中i,j
对应于x,y
的索引np.unique
? In the example this would be something like 在示例中,这类似于
Z = [[ 2, 0, 3],
['', '', 1],
[ 1, 0, '']]
where ''
might as well be np.nan
. 其中
''
最好是np.nan
。 This does somehow sound like an inverse np.meshgrid
, and I could hack up my own implementation, but is there no pre-existing solution? 这听起来像是一个反向
np.meshgrid
,我可以破解自己的实现,但是没有预先存在的解决方案吗?
I tried the suggestions here , but they assume a complete grid. 我在这里尝试了这些建议,但它们假定使用完整的网格。 Another solution sounds nice but interpolates the missing points, which is not what I want.
另一个解决方案听起来不错,但可以插补缺失的点,这不是我想要的。
One approach would be - 一种方法是-
m,n = np.max(x)+1, np.max(y)+1
out = np.full((m,n), np.nan)
out[x,y] = z
Sample run - 样品运行-
In [213]: x = [4,0,2,2,1,0]
...: y = [0,0,0,1,2,5]
...: z = [0,2,3,1,0,1]
...:
In [214]: m,n = np.max(x)+1, np.max(y)+1
...: out = np.full((m,n), np.nan)
...: out[x,y] = z
...:
In [215]: out
Out[215]:
array([[ 2., nan, nan, nan, nan, 1.],
[ nan, nan, 0., nan, nan, nan],
[ 3., 1., nan, nan, nan, nan],
[ nan, nan, nan, nan, nan, nan],
[ 0., nan, nan, nan, nan, nan]])
For floating point values, we could use np.unique(..return_inverse)
to give each of the X's and Y's unique int IDs, which could be used as row and column indices for indexing into output array - 对于浮点值,我们可以使用
np.unique(..return_inverse)
给出X和Y的每个唯一的int ID,这些ID可用作行索引和列索引,以索引到输出数组中-
x_arr = np.unique(x, return_inverse=1)[1]
y_arr = np.unique(y, return_inverse=1)[1]
m,n = np.max(x_arr)+1, np.max(y_arr)+1
out = np.full((m,n), np.nan)
out[x_arr,y_arr] = z
Sample run - 样品运行-
In [259]: x = [1, 0,.2,.2, 1, 0]
...: y = [0, 0, 0, 1,.2,.2]
...: z = [0, 2, 3, 1, 0, 1]
...:
In [260]: x_arr = np.unique(x, return_inverse=1)[1]
...: y_arr = np.unique(y, return_inverse=1)[1]
...:
...: m,n = np.max(x_arr)+1, np.max(y_arr)+1
...: out = np.full((m,n), np.nan)
...: out[x_arr,y_arr] = z
...:
In [261]: out
Out[261]:
array([[ 2., 1., nan],
[ 3., nan, 1.],
[ 0., 0., nan]])
Based on Divakar's answer , but also working for non-index x,y
s: 基于Divakar的答案 ,但也适用于非索引
x,y
s:
ux, xi = np.unique(x, return_inverse=1)
uy, yi = np.unique(y, return_inverse=1)
X, Y = np.meshgrid(ux, uy)
Z = np.full(X.shape, np.nan)
Z[xi, yi] = z
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