删除列表中包含的numpy行? [英] Remove numpy rows contained in a list?

问题描述

我有一个numpy数组和一个列表.我要删除列表中包含的行.

I have a numpy array and a list. I want to remove the rows contained in the list.

a = np.zeros((3, 2))
a[0, :] = [1, 2]
l = [(1, 2), (3, 4)]

目前，我尝试通过创建一组a行来做到这一点，然后排除从l创建的set，例如:

Currently I try to do this by making a set of a's rows, then exclude the set created from l, something like:

sa = set(map(tuple, a))
sl = set(l)
np.array(list(sa - sl))

或更简单地

sl = set(l)
np.array([row for row in list(map(tuple, a)) if row not in sl]

当每一行都很短时，这些效果很好.

These work pretty well when each row is short.

有更快的方法吗?我需要优化速度.

Is there a faster way? I need to optimize for speed.

推荐答案

方法#1:这是views的视图(将每行视为具有扩展dtype的元素)-

Approach #1 : Here's one with views (viewing each row as an element each with extended dtype) -

# https://stackoverflow.com/a/45313353/ @Divakar
def view1D(a, b): # a, b are arrays
    a = np.ascontiguousarray(a)
    b = np.ascontiguousarray(b)
    void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
    return a.view(void_dt).ravel(),  b.view(void_dt).ravel()

a1D,l1D = view1D(a,l)
out = a[np.in1d(a1D,l1D,invert=True)]

如果仅需要像set一样在输出中具有唯一的行，请对获得的输出使用np.unique-

If you need to have unique rows only in the output as with set, use np.unique on the output obtained -

np.unique(out,axis=0)

样品运行输出-

In [72]: a
Out[72]: 
array([[1, 2],
       [0, 0],
       [0, 0]])

In [73]: l
Out[73]: [(1, 2), (3, 4)]

In [74]: out
Out[74]: 
array([[0, 0],
       [0, 0]])
In [75]: np.unique(out,axis=0)
Out[75]: array([[0, 0]])

方法2::降维原理相同，这是特定于int dtype数据的矩阵乘法-

Approach #2 : With the same philosophy of reducing dimensionality, here's with matrix-multiplication specific to int dtype data -

l = np.asarray(l)
shp = np.maximum(a.max(0)+1,l.max(0)+1)
s = np.r_[shp[::-1].cumprod()[::-1][1:],1]
l1D = l.dot(s)
a1D = a.dot(s)
l1Ds = np.sort(l1D)
out = a[l1D[np.searchsorted(l1Ds,a1D)] != a1D]

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