[英]Compare two numpy arrays by first Column and create a third numpy array by concatenating two arrays
我有兩個2d numpy數組,用於繪制模擬結果。
數組a
和b
的第一列包含時間間隔,第二列包含要繪制的數據。 這兩個數組具有不同的形狀a(500,2)
b(600,2)
。 我想第一列這兩個numpy的陣列比較和創建上的第一列中找到相匹配的第三組a
。 如果找不到匹配項,則將0添加到第三列。
有什么numpy技巧可以做到這一點嗎?
例如:
a=[[0.002,0.998],
[0.004,0.997],
[0.006,0.996],
[0.008,0.995],
[0.010,0.993]]
b= [[0.002,0.666],
[0.004,0.665],
[0.0041,0.664],
[0.0042,0.664],
[0.0043,0.664],
[0.0044,0.663],
[0.0045,0.663],
[0.0005,0.663],
[0.006,0.663],
[0.0061,0.662],
[0.008,0.661]]
預期產量
c= [[0.002,0.998,0.666],
[0.004,0.997,0.665],
[0.006,0.996,0.663],
[0.008,0.995,0.661],
[0.010,0.993, 0 ]]
我可以很快將解決方案視為
import numpy as np
a = np.array([[0.002, 0.998],
[0.004, 0.997],
[0.006, 0.996],
[0.008, 0.995],
[0.010, 0.993]])
b = np.array([[0.002, 0.666],
[0.004, 0.665],
[0.0041, 0.664],
[0.0042, 0.664],
[0.0043, 0.664],
[0.0044, 0.663],
[0.0045, 0.663],
[0.0005, 0.663],
[0.0006, 0.663],
[0.00061, 0.662],
[0.0008, 0.661]])
c = []
for row in a:
index = np.where(b[:,0] == row[0])[0]
if np.size(index) != 0:
c.append([row[0], row[1], b[index[0], 1]])
else:
c.append([row[0], row[1], 0])
print c
如上面的評論所指出,似乎存在數據輸入錯誤
import numpy as np
i = np.intersect1d(a[:,0], b[:,0])
overlap = np.vstack([i, a[np.in1d(a[:,0], i), 1], b[np.in1d(b[:,0], i), 1]]).T
underlap = np.setdiff1d(a[:,0], b[:,0])
underlap = np.vstack([underlap, a[np.in1d(a[:,0], underlap), 1], underlap*0]).T
fast_c = np.vstack([overlap, underlap])
通過使用intersect1d獲取 a
和b
的第一列的交集 ,然后使用in1d與第二列的交集進行交叉引用,可以實現此目的。
vstack
垂直堆疊輸入的元素,並且需要轉置以獲得正確的尺寸(非常快速的操作)。
然后找到在時間a
不在b
使用setdiff1d ,並通過把0在第三列完成的結果。
打印出來
array([[ 0.002, 0.998, 0.666],
[ 0.004, 0.997, 0.665],
[ 0.006, 0.996, 0. ],
[ 0.008, 0.995, 0. ],
[ 0.01 , 0.993, 0. ]])
以下內容適用於numpy數組和簡單的python列表。
c = [[*x, y[1]] for x in a for y in b if x[0] == y[0]]
d = [[*x, 0] for x in a if x[0] not in [y[0] for y in b]]
c.extend(d)
比我勇敢的人可以嘗試做這一行。
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