Python—如何通过保持K个最大值来减少每行或对称矩阵的条目数

我有一个对称的相似矩阵，我想在每一行中只保留k个最大值

这里有一些代码完全符合我的要求，但我想知道是否有更好的方法。尤其是展平/重塑有点笨拙。提前谢谢

请注意，nrows（如下）必须扩展到数万个

from scipy.spatial.distance import pdist, squareform
random.seed(1)
nrows = 4
a = (random.rand(nrows,nrows))

# Generate a symmetric similarity matrix
s = 1-squareform( pdist( a, 'cosine' ) )
print "Start with:\n", s

# Generate the sorted indices
ss = argsort(s.view(np.ndarray), axis=1)[:,::-1]
s2 = ss + (arange(ss.shape[0])*ss.shape[1])[:,None]

# Zero-out after k-largest-value entries in each row
k = 3 # Number of top-values to keep, per row
s = s.flatten()
s[s2[:,k:].flatten()] = 0
print "Desired output:\n", s.reshape(nrows,nrows)

给出：

Start with:
[[ 1.          0.61103296  0.82177072  0.92487807]
 [ 0.61103296  1.          0.94246304  0.7212526 ]
 [ 0.82177072  0.94246304  1.          0.87247418]
 [ 0.92487807  0.7212526   0.87247418  1.        ]]
Desired output:
[[ 1.          0.          0.82177072  0.92487807]
 [ 0.          1.          0.94246304  0.7212526 ]
 [ 0.          0.94246304  1.          0.87247418]
 [ 0.92487807  0.          0.87247418  1.        ]]

---

这不是很大的改进，但是为了避免扁平化和重塑，您可以使用

np.put

：

# Generate the sorted indices
ss = np.argsort(s.view(np.ndarray), axis=1)[:,::-1]
ss += (np.arange(ss.shape[0])*ss.shape[1])[:,None] #Add in place, probably trivial improvement

k=3
np.put(s,ss[:,k:],0) #or s.flat[ss[:,k:]]=0
print s

[[ 1.          0.          0.82177072  0.92487807]
 [ 0.          1.          0.94246304  0.7212526 ]
 [ 0.          0.94246304  1.          0.87247418]
 [ 0.92487807  0.          0.87247418  1.        ]]

---

如果您发现自己正在将索引的长列表生成到一个数组中，那么很有可能使用布尔矩阵以更优雅的方式来解决这个问题。就你而言：

a = np.random.rand(5, 5)
a = a + a.T # make it symmetrical

sort_idx = np.argsort(np.argsort(a, axis=1), axis=1)
k = 3 # values to keep
# if you want a copy of the original
mask = (sort_idx >= a.shape[1] - k) # positions we want to keep
b = np.zeros_like(a)
b[mask] = a[mask]

# if you wantrd to do the operation in-place
# mask = (sort_idx < a.shape[1] - k) # positions we want to zero
# a[mask] = 0

>>> a
array([[ 1.87816548,  0.86562424,  1.94171234,  0.96565312,  0.53451029],
       [ 0.86562424,  1.13762348,  1.48565754,  0.78031763,  0.51448499],
       [ 1.94171234,  1.48565754,  1.39960519,  0.57456214,  1.32608456],
       [ 0.96565312,  0.78031763,  0.57456214,  1.56469221,  0.74632264],
       [ 0.53451029,  0.51448499,  1.32608456,  0.74632264,  0.55378676]])
>>> b
array([[ 1.87816548,  0.        ,  1.94171234,  0.96565312,  0.        ],
       [ 0.86562424,  1.13762348,  1.48565754,  0.        ,  0.        ],
       [ 1.94171234,  1.48565754,  1.39960519,  0.        ,  0.        ],
       [ 0.96565312,  0.78031763,  0.        ,  1.56469221,  0.        ],
       [ 0.        ,  0.        ,  1.32608456,  0.74632264,  0.55378676]])

a=np.random.rand（5,5）
a=a+a.T#使其对称
sort_idx=np.argsort（np.argsort（a，轴=1），轴=1）
k=3#要保持的值
#如果你想要原件的复印件
mask=（sort_idx>=a.shape[1]-k）#我们要保留的位置
b=np.类零（a）
b[遮罩]=a[遮罩]
#如果你想在适当的地方做手术
#mask=（sort_idx>>a
数组（[[1.87816548,0.86562424,1.94171234,0.96565312,0.53451029]，
[ 0.86562424,  1.13762348,  1.48565754,  0.78031763,  0.51448499],
[ 1.94171234,  1.48565754,  1.39960519,  0.57456214,  1.32608456],
[ 0.96565312,  0.78031763,  0.57456214,  1.56469221,  0.74632264],
[ 0.53451029,  0.51448499,  1.32608456,  0.74632264,  0.55378676]])
>>>b
数组（[[1.87816548,0,1.94171234,0.96565312,0.]，
[ 0.86562424,  1.13762348,  1.48565754,  0.        ,  0.        ],
[ 1.94171234,  1.48565754,  1.39960519,  0.        ,  0.        ],
[ 0.96565312,  0.78031763,  0.        ,  1.56469221,  0.        ],
[ 0.        ,  0.        ,  1.32608456,  0.74632264,  0.55378676]])

正是我想要的。谢谢奥菲恩！您的第一行隐藏了最大值（1.909…），它应该保留每行的三个（或“k”）最大值，对吗？@zbinsd是的，您完全正确。通过再次运行

argsort

可以很容易地解决此问题，请参见我的编辑。我还更改了索引条件，以保留最大值，而不是最小值。