我正在使用np.digitize将Python中的2d数组(x by y)合并到其x值的区间(在"bins"中给出):
elements_to_bins = digitize(vals, bins)
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其中"vals"是一个二维数组,即:
vals = array([[1, v1], [2, v2], ...]).
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elements_to_bins只是说每个元素落入哪个bin.我当时想要做的是得到一个列表,其长度是"箱子"中的箱数,每个元素返回落入该箱的"val"的y维度.我现在这样做:
points_by_bins = []
for curr_bin in range(min(elements_to_bins), max(elements_to_bins) + 1):
curr_indx = where(elements_to_bins == curr_bin)[0]
curr_bin_vals = vals[:, curr_indx]
points_by_bins.append(curr_bin_vals)
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有没有更优雅/更简单的方法来做到这一点?我只需要列出每个bin中的y值列表.
谢谢.
如果我正确理解你的问题:
\n\nvals = array([[1, 10], [1, 11], [2, 20], [2, 21], [2, 22]]) # Example\n\n(x, y) = vals.T # Shortcut\nbin_limits = range(min(x)+1, max(x)+2) # Other limits could be chosen\npoints_by_bin = [ []\xc2\xa0for _ in bin_limits ] # Final result\nfor (bin_num, y_value) in zip(searchsorted(bin_limits, x, "right"), y): # digitize() finds the correct bin number\n points_by_bin[bin_num].append(y_value)\n\nprint points_by_bin # [[10, 11], [20, 21, 22]]\n
Run Code Online (Sandbox Code Playgroud)\n\nNumpy 的快速数组运算searchsorted()
用于实现最大效率。然后将值一一相加(由于最终结果不是矩形数组,因此 Numpy 对此无能为力)。此解决方案应该比循环中的多次调用更快 ,循环中的多次调用迫使 Numpy多次where()
重新读取同一数组。