我有兩個包含 XYZ 點 A 和 B 的二維陣列。
陣列 A 的形狀為 (796704, 3)，是我原來的點云。除了 (0, 0, 0) 之外，每個點都是唯一的，但這些都無關緊要：
A = [[x_1, y_1, z_1],
[x_2, y_2, z_2],
[x_3, y_3, z_3],
[x_4, y_4, z_4],
[x_5, y_5, z_5],
...]
陣列 B 的形狀為 (N, 4)，是 A 的裁剪版本 (N<796704)。
其余點沒有變化，仍然與 A 中的對應點相同
。第四列包含每個點的分割值。
B 的行順序是完全隨機的，不再與 A 匹配。
B = [[x_4, y_4, z_4, 5],
[x_2, y_2, z_2, 12],
[x_6, y_6, z_6, 5],
[x_7, y_7, z_7, 3],
[x_9, y_9, z_9, 3]]
我需要重新排序 B 的行，以便它們與 A 的行匹配相同的點，并用零行填充空白：
B = [[0.0, 0.0, 0.0, 0],
[x_2, y_2, z_2, 12],
[0.0, 0.0, 0.0, 0],
[x_4, y_4, z_4, 5],
[0.0, 0.0, 0.0, 0],
[x_6, y_6, z_6, 5],
[x_7, y_7, z_7, 3],
[0.0, 0.0, 0.0, 0],
[x_9, y_9, z_9, 3],
[0.0, 0.0, 0.0, 0],
[0.0, 0.0, 0.0, 0],
[0.0, 0.0, 0.0, 0]
...]
最后 B 應該具有形狀 (796704, 4)。
我嘗試使用 numpy_indexed 包，就像在這個非常相似的問題中提出的那樣，但這里的問題是 B 不包含 A 的所有點：
import numpy_indexed as npi
B[npi.indices(B[:, :-1], A)]
我對 numpy 不熟悉，我唯一的解決方案是一個 for 回圈，但這對于我的應用程式來說太慢了。有沒有解決這個問題的某種快速方法？
uj5u.com熱心網友回復：
熊貓 => 重新索引：
import pandas as pd
import numpy as np
A = np.array([[8, 7, 4],
[0, 7, 7],
[4, 7, 0],
[5, 5, 8],
[8, 7, 5]])
B = np.array([[8, 7, 4, 2],
[4, 7, 0, 5],
[8, 7, 5, 6]])
df_B = (pd.DataFrame(B, columns=["x", "y", "z", "seg"])
.set_index(["x", "y", "z"])
.reindex(list(map(tuple, A)))
.reset_index())
df_B.loc[df_B.seg.isna()] = 0
B = df_B.values
print(B)
結果：
array([[8., 7., 4., 2.],
[0., 0., 0., 0.],
[4., 7., 0., 5.],
[0., 0., 0., 0.],
[8., 7., 5., 6.]])
uj5u.com熱心網友回復：
僅使用 numpy 解決您的問題：
情況1
您只使用數字：
import numpy as np
A = np.array([[1, 1, 1],
[2, 2, 2],
[3, 3, 3],
[4, 4, 4],
[5, 5, 5],
[6, 6, 6],
[7, 7, 7],
[8, 8, 8],
[9, 9, 9],
[10,10, 10]
])
B = np.array([[4, 4, 4, 5],
[2, 2, 2, 12],
[6, 6, 6, 5],
[7, 7, 7, 3],
[9, 9, 9, 3]])
c = np.insert(A, 3, 0, axis = 1)
d = np.vstack((B,c[np.in1d(c[:,0],B[:,0], invert=True)]*0))
print(d)
Out:
[[ 4 4 4 5]
[ 2 2 2 12]
[ 6 6 6 5]
[ 7 7 7 3]
[ 9 9 9 3]
[ 0 0 0 0] # previously 1, 1, 1, 0
[ 0 0 0 0] # previously 3, 3, 3, 0
[ 0 0 0 0] # previously 5, 5, 5, 0
[ 0 0 0 0] # previously 8, 8, 8, 0
[ 0 0 0 0]] # previously 10, 10, 10, 0
解釋：
1oc將是一個A帶有新欄位的副本0：
c = np.insert(A, 3, 0, axis = 1)
如果我現在列印c，我會得到這個：
[[ 1 1 1 0]
[ 2 2 2 0]
[ 3 3 3 0]
[ 4 4 4 0]
[ 5 5 5 0]
[ 6 6 6 0]
[ 7 7 7 0]
[ 8 8 8 0]
[ 9 9 9 0]
[10 10 10 0]]
2o 您創建一個新陣列B，其中的部分c不B乘以0。
d = np.vstack((B,c[np.in1d(c[:,0],B[:,0], invert=True)]*0))
2.1 np.vstack((B,_)) Here I removed the c just to be more easy to you to see the args that vstack receive. You have a tuple with the two arrays that you want to concatenate.
2.2 c[np.in1d(c[:,0],B[:,0], invert=True)]*0 Instead of passing all the c a pass c selecting np.in1d(c[:,0],B[:,0], invert=True) of c and multiplying it by 0.
2.3 np.in1d(c[:,0],B[:,0], invert=True) If I do np.in1d(c[:,0],B[:,0]) I get a boolean array telling me which x_n of c also exists in B, if I set invert=True i'll get which x_n of c does NOT exists in B. (Another way to to that invertion is by using the tilde operator ~, so ~np.in1d(c[:,0],B[:,0]) == np.in1d(c[:,0],B[:,0], invert=True))
Since each point is unique with the exception of the 0,0,0,0 ones, when I do c[np.in1d(c[:,0],B[:,0], invert=True)] I get:
array([[ 1, 1, 1, 0],
[ 3, 3, 3, 0],
[ 5, 5, 5, 0],
[ 8, 8, 8, 0],
[10, 10, 10, 0]])
if I multiply by 0 I get:
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]])
So in np.vstack((B,c[np.in1d(c[:,0],B[:,0], invert=True)]*0)) I concatenate the B and the c.
Being the B this:
array([[ 4, 4, 4, 5],
[ 2, 2, 2, 12],
[ 6, 6, 6, 5],
[ 7, 7, 7, 3],
[ 9, 9, 9, 3]])
and c the array of 0's above. The result at the end is:
array([[ 4, 4, 4, 5],
[ 2, 2, 2, 12],
[ 6, 6, 6, 5],
[ 7, 7, 7, 3],
[ 9, 9, 9, 3],
[ 0, 0, 0, 0],
[ 0, 0, 0, 0],
[ 0, 0, 0, 0],
[ 0, 0, 0, 0],
[ 0, 0, 0, 0]])
Case 2
If you are working with strings and numbers you can do that way:
import numpy as np
A = np.array([['x_1', 'y_1', 'z_1'],
['x_2', 'y_2', 'z_2'],
['x_3', 'y_3', 'z_3'],
['x_4', 'y_4', 'z_4'],
['x_5', 'y_5', 'z_5'],
['x_6', 'y_6', 'z_6'],
['x_7', 'y_7', 'z_7'],
['x_8', 'y_8', 'z_8'],
['x_9', 'y_9', 'z_9'],
['x_10', 'y_10', 'z_10']
])
B = np.array([['x_4', 'y_4', 'z_4', 5],
['x_2', 'y_2', 'z_2', 12],
['x_6', 'y_6', 'z_6', 5],
['x_7', 'y_7', 'z_7', 3],
['x_9', 'y_9', 'z_9', 3]])
c = np.insert(A, 3, 0, axis = 1)
c[np.in1d(c[:,0],B[:,0], invert=True)] = 0
d = np.vstack((B,c[np.in1d(c[:,0],B[:,0], invert=True)]))
print(d)
Out:
[['x_4' 'y_4' 'z_4' '5']
['x_2' 'y_2' 'z_2' '12']
['x_6' 'y_6' 'z_6' '5']
['x_7' 'y_7' 'z_7' '3']
['x_9' 'y_9' 'z_9' '3']
['0' '0' '0' '0']
['0' '0' '0' '0']
['0' '0' '0' '0']
['0' '0' '0' '0']
['0' '0' '0' '0']]
Explanation:
1o c will be a copy of A with a new field with a 0:
c = np.insert(A, 3, 0, axis = 1)
If I print c right now I will get this:
[['x_1' 'y_1' 'z_1' '0']
['x_2' 'y_2' 'z_2' '0']
['x_3' 'y_3' 'z_3' '0']
['x_4' 'y_4' 'z_4' '0']
['x_5' 'y_5' 'z_5' '0']
['x_6' 'y_6' 'z_6' '0']
['x_7' 'y_7' 'z_7' '0']
['x_8' 'y_8' 'z_8' '0']
['x_9' 'y_9' 'z_9' '0']
['x_10' 'y_10' 'z_10' '0']]
2o At the fields of c that don't exist in B, i'll set as 0
c[np.in1d(c[:,0],B[:,0], invert=True)] = 0
3o d will be B the c part that was set as 0
d = np.vstack((B,c[np.in1d(c[:,0],B[:,0], invert=True)]))
由于在這種情況下，您正在處理同一陣列中的字串和數字，因此您不能只將 all 乘以0. d所以你需要設定cas的欄位，0然后選擇0's 的欄位。
有用的鏈接：
我的答案基于.
波浪號運算子。
uj5u.com熱心網友回復：
我設法通過使用我在問題中提到的numpy_indexed包解決了這個問題。
解決方案：
A = np.array([[8, 7, 4],
[0, 7, 7],
[4, 3, 0],
[5, 5, 8],
[3, 9, 5]])
B = np.array([[3, 9, 5, 6],
[8, 7, 4, 2],
[4, 3, 0, 5]])
# Create a new, zero-filled, array C with length of A
C = np.zeros((A.shape[0], 4))
# Insert B at the beginning of C
C[:B.shape[0], :B.shape[1]] = B
print(C)
Out:
[[3, 9, 5, 6],
[8, 7, 4, 2],
[4, 3, 0, 5],
[0, 0, 0, 0],
[0, 0, 0, 0]]
# Using the numpy_indexed package reorder the rows.
# The last index of C is used as a fill value in case
# a row wasn't found in A thus filling the gaps with [0,0,0,0]
import numpy_indexed as npi
D = C[npi.indices(C[:, :-1], A, missing=-1)]
print(D)
Out:
[[8, 7, 4, 2],
[0, 0, 0, 0],
[4, 3, 0, 5],
[0, 0, 0, 0],
[3, 9, 5, 6]]
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