此备忘录是否正常工作？(Is this memoization working properly?)
我一直在Haskell上努力解决Project Euler＃14一段时间，但出于某种原因，我无法让它工作。 我刚才使用Groovy解决了这个问题，我想我在这里使用的方法基本相同。 然而，程序运行速度非常慢，即使只是找到前10,000个长度，我现在真的迷失了原因。 我认为我正在使用memoization，但即使GHCI中的数据集很少，我也会耗尽内存。
这是我到目前为止所提出的。
collatz = (map collatz' [0..] !!) where collatz' n | n == 1 = 1 | n `mod` 2 == 0 = 1 + collatz (n `div` 2) | otherwise = 1 + collatz (3 * n + 1)我将运行
map collatz [1..1000000]来获得问题的答案，但map collatz [1..10000]给我一个内存不足的错误，并且还需要几秒钟才能完成运行。如果有人能给我一些关于这个程序的问题的见解，那就太好了！ 我尝试了很多东西，我只是陷入困境，需要一只手。
谢谢！
I'm have been working on solving Project Euler #14 for a while now in Haskell, but for some reason, I'm unable to get it working. I solved the problem using Groovy a while ago, and I think I'm using basically the same method here. However, the program runs incredibly slow even just finding the first 10,000 lengths, and I'm really lost now as to why. I think I'm using memoization right, but I'm running out of memory even with smallish data sets in GHCI.
Here's what I've come up with so far.
collatz = (map collatz' [0..] !!) where collatz' n | n == 1 = 1 | n `mod` 2 == 0 = 1 + collatz (n `div` 2) | otherwise = 1 + collatz (3 * n + 1)I'd be running
map collatz [1..1000000]to get the answer to the problem, butmap collatz [1..10000]gives me an out of memory error, and also takes a good few seconds to finish running.If anyone could give me some insights as to what the problem with this program is, that would be great! I've tried a lot of things and I'm just stuck and need a hand.
Thanks!
原文：https://stackoverflow.com/questions/11849664
最满意答案
row_index是每列最大的索引，50列，50个索引。
col_index是每行最大的索引，35个值。在这两个数组中配对值是没有意义的。 是的，其中一对将是整个数组最大值的索引，但你不能仅仅通过查看它们来判断。
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In [573]: x Out[573]: array([[15, 8, 6, 3, 4, 5], [22, 1, 13, 10, 18, 7], [21, 14, 20, 16, 9, 2], [11, 12, 23, 19, 17, 0]]) In [574]: i0=np.argmax(x,axis=0) In [575]: i1=np.argmax(x,axis=1) In [576]: i0 Out[576]: array([1, 2, 3, 3, 1, 1], dtype=int32) In [577]: i1 Out[577]: array([0, 0, 0, 2], dtype=int32)要获取最大值，我们必须将这些索引与正确的
np.arange形状组合在一起。In [578]: x[i0,np.arange(6)] Out[578]: array([22, 14, 23, 19, 18, 7]) In [579]: x[np.arange(4),i1] Out[579]: array([15, 22, 21, 23])
23存在于两者中，但不存在于相同的索引处In [580]: np.argmax(x) Out[580]: 20 # location in the flattened version of x In [581]: x.flat[20] Out[581]: 23 In [582]: x[i0[2], i1[3]] # (3, 2) Out[582]: 23要将平面索引转换为2d，请使用：
In [583]: np.unravel_index(20,x.shape) Out[583]: (3, 2) In [584]: x[3,2] Out[584]: 23
row_indexis the index of the maximum for each column, 50 columns, 50 indices.
col_indexis the index of the maximum for each rows, 35 values.It doesn't make sense to pair up values in those two arrays. Yes, one of those pairs will be the index of the whole-array max, but you can't tell just by looking them.
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In [573]: x Out[573]: array([[15, 8, 6, 3, 4, 5], [22, 1, 13, 10, 18, 7], [21, 14, 20, 16, 9, 2], [11, 12, 23, 19, 17, 0]]) In [574]: i0=np.argmax(x,axis=0) In [575]: i1=np.argmax(x,axis=1) In [576]: i0 Out[576]: array([1, 2, 3, 3, 1, 1], dtype=int32) In [577]: i1 Out[577]: array([0, 0, 0, 2], dtype=int32)To fetch the max values, we have to combine these indices with the correct shape of
np.arange.In [578]: x[i0,np.arange(6)] Out[578]: array([22, 14, 23, 19, 18, 7]) In [579]: x[np.arange(4),i1] Out[579]: array([15, 22, 21, 23])
23is present in both, but not at the same indexIn [580]: np.argmax(x) Out[580]: 20 # location in the flattened version of x In [581]: x.flat[20] Out[581]: 23 In [582]: x[i0[2], i1[3]] # (3, 2) Out[582]: 23To convert the flat index to 2d, use:
In [583]: np.unravel_index(20,x.shape) Out[583]: (3, 2) In [584]: x[3,2] Out[584]: 23
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