事实证明，在这种情况下，纯Python循环比NumPy索引（或对np.where的调用）要快得多。
考虑以下替代方案：
import numpy as npimport collectionsimport itertools as ITshape = (2600,5200)# shape = (26,52)emiss_data = np.random.random(shape)obj_data = np.random.random_integers(1, 800, size=shape)UNIQ_IDS = np.unique(obj_data)def using_where(): max = np.max where = np.where MAX_EMISS = [max(emiss_data[where(obj_data == i)]) for i in UNIQ_IDS] return MAX_EMISSdef using_index(): max = np.max MAX_EMISS = [max(emiss_data[obj_data == i]) for i in UNIQ_IDS] return MAX_EMISSdef using_max(): MAX_EMISS = [(emiss_data[obj_data == i]).max() for i in UNIQ_IDS] return MAX_EMISSdef using_loop(): result = collections.defaultdict(list) for val, idx in IT.izip(emiss_data.ravel(), obj_data.ravel()): result[idx].append(val) return [max(result[idx]) for idx in UNIQ_IDS]def using_sort(): uind = np.digitize(obj_data.ravel(), UNIQ_IDS) - 1 vals = uind.argsort() count = np.bincount(uind) start = 0 end = 0 out = np.empty(count.shape[0]) for ind, x in np.ndenumerate(count): end += x out[ind] = np.max(np.take(emiss_data, vals[start:end])) start += x return outdef using_split(): uind = np.digitize(obj_data.ravel(), UNIQ_IDS) - 1 vals = uind.argsort() count = np.bincount(uind) return [np.take(emiss_data, item).max() for item in np.split(vals, count.cumsum())[:-1]]for func in (using_index, using_max, using_loop, using_sort, using_split): assert using_where() == func()
以下是基准测试shape = (2600,5200)：
In [57]: %timeit using_loop()1 loops, best of 3: 9.15 s per loopIn [90]: %timeit using_sort()1 loops, best of 3: 9.33 s per loopIn [91]: %timeit using_split()1 loops, best of 3: 9.33 s per loopIn [61]: %timeit using_index()1 loops, best of 3: 63.2 s per loopIn [62]: %timeit using_max()1 loops, best of 3: 64.4 s per loopIn [58]: %timeit using_where()1 loops, best of 3: 112 s per loop
因此，using_loop（纯Python）的运行速度比快11倍以上using_where。
我不完全确定为什么纯Python比NumPy更快。我的猜测是纯Python版本一次通过两个数组（是，是双关语）。它利用了这样一个事实，即尽管进行了所有花哨的索引编制，但实际上我们只想访问每个值一次。因此，它避免了必须确定每个值所属的组的问题emiss_data。但这只是模糊的推测。我不知道在进行基准测试之前会更快。
