我有以下 numpy 數(shù)組：theta_array =array([[ 1, 10], [ 1, 11], [ 1, 12], [ 1, 13], [ 1, 14], [ 2, 10], [ 2, 11], [ 2, 12], [ 2, 13], [ 2, 14], [ 3, 10], [ 3, 11], [ 3, 12], [ 3, 13], [ 3, 14], [ 4, 10], [ 4, 11], [ 4, 12], [ 4, 13], [ 4, 14]])XY_array = array([[ 44.0394952 , 505.81099922], [ 61.03882938, 515.97253226], [ 26.69851841, 525.18083012], [ 46.78487831, 533.42309602], [ 45.77188401, 545.42988355], [ 81.12969132, 554.78767379], [ 54.178463 , 565.8716283 ], [ 41.58952084, 574.76827133], [ 85.24956815, 585.1355127 ], [ 80.73726733, 595.49446033], [ 22.70625059, 605.59017175], [ 40.66810604, 615.26308629], [ 47.16694695, 624.39222332], [ 48.72499541, 633.19846364], [ 50.68589921, 643.72334885], [ 38.42731134, 654.68595883], [ 47.39519707, 666.28232866], [ 58.07767155, 673.9572227 ], [ 72.11393347, 683.68307373], [ 53.70872932, 694.65509894], [ 82.08237952, 704.5868817 ], [ 46.64069738, 715.18427515], [ 40.46032478, 723.91308011], [ 75.69090892, 733.69595658], [120.61447884, 745.31322786], [ 60.17764744, 754.89747186], [ 87.15961973, 766.24040447], [ 82.93872713, 773.01518252], [ 93.56688906, 785.60640153], [ 70.0474047 , 793.81792947], [104.3613818 , 805.40234676], [108.39253837, 814.75002114], [ 78.97643673, 824.95386427], [ 85.69096895, 834.44797862], [ 53.07112931, 844.39555058], [111.49875807, 855.660508 ], [ 70.88824958, 865.53417489], [ 79.55499469, 875.31303945], [ 60.86941464, 885.85235946], [101.06017712, 896.69986636], [ 74.55823544, 905.87417231], [113.24705653, 915.19350121], [ 94.21920882, 925.87933273],我試圖將兩者結(jié)合起來，因此對于 theta_array 的每個組合，我從 XY_array 獲取所有組合。在 numpy 中進行這種組合/聚合的方式是什么？
3 回答
浮云間
TA貢獻1829條經(jīng)驗 獲得超4個贊
meshgrid您可以在使用with后對“列”重新排序[:,[0,2,1,3]]，如果由于列數(shù)較多而需要使列表動態(tài)化，那么您可以看到我答案的結(jié)尾：
np.array(np.meshgrid(theta_array, XY_array)).T.reshape(-1,4)[:,[0,2,1,3]]
輸出：
array([[ 1. , 1. , 44.0394952 , 505.81099922]],
[[ 1. , 1. , 61.03882938, 515.97253226]],
[[ 1. , 1. , 26.69851841, 525.18083012]],
...,
[[ 14. , 14. , 73.86032436, 973.91032818]],
[[ 14. , 14. , 103.96923524, 984.24366761]],
[[ 14. , 14. , 93.20663129, 995.44618851]])
如果您有很多列，您可以動態(tài)創(chuàng)建此列表：[0,2,1,3]使用列表理解。例如：
n = new_arr.shape[1]*2
lst = [x for x in range(n) if x % 2 == 0]
[lst.append(z) for z in [y for y in range(n) if y % 2 == 1]]
lst
[0, 2, 4, 6, 1, 3, 5, 7]
然后，您可以重寫為：
np.array(np.meshgrid(theta_array, XY_array)).T.reshape(-1,4)[:,lst]
哆啦的時光機
TA貢獻1779條經(jīng)驗 獲得超6個贊
您可以使用itertools.product：
out = np.array([*product(theta_array, XY_array)])
out = out.reshape(out.shape[0],-1)
輸出：
array([[ 1. , 10. , 44.0394952 , 505.81099922],
[ 1. , 10. , 61.03882938, 515.97253226],
[ 1. , 10. , 26.69851841, 525.18083012],
...,
[ 4. , 14. , 73.86032436, 973.91032818],
[ 4. , 14. , 103.96923524, 984.24366761],
[ 4. , 14. , 93.20663129, 995.44618851]])
也就是說，這看起來非常像XY 問題。你想用這個數(shù)組做什么？
鴻蒙傳說
TA貢獻1865條經(jīng)驗 獲得超7個贊
正如此處的側(cè)面/補充參考一樣，我們對兩種解決方案的執(zhí)行時間進行了比較。完成此特定操作itertools所需的時間比同等操作多 10 倍numpy。
%%time
for i in range(1000):
z = np.array(np.meshgrid(theta_array, XY_array)).T.reshape(-1,4)[:,[0,2,1,3]]
CPU times: user 299 ms, sys: 0 ns, total: 299 ms
Wall time: 328 ms
%%time
for i in range(1000):
z = np.array([*product(theta_array, XY_array)])
z = z.reshape(z.shape[0],-1)
CPU times: user 2.79 s, sys: 474 μs, total: 2.79 s
Wall time: 2.84 s
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