2015-05-19 67 views
1

我有一個numpy矩陣A,我需要一個函數來計算(A [i,j] /第i列中所有元素的和) - A [i, J] /在第j行 所有元素的總和,它爲A.每個元素如何在Python中編寫numpy矩陣的函數

我已經試着寫沿着線的東西:

a = A.sum(axis=0) 
b = A.sum(axis=1) 

for i in A: 
    for j in i: 
     f = A[i,j]/a(i) - A[i,j]/b(j) 
+0

你能否多說一點你試過的東西? – senderle

+0

是的,我試圖寫不服像: 一個= A.sum(軸= 0) B = A.sum(軸= 1) 對於i在一個: 對於j在I: F = A [1 ,j]/a(i)-A [i,j]/b(j) – Polly

回答

0

我希望這可以幫助:

def f(i,j,A): 
    return A[i,j] * (1./np.sum(A[i,]) ) - (1./np.sum(A[:,j]) ) 
2

你應該尋找av ectorized解決方案不是通過

A/A.sum(axis=0)[np.newaxis,:] - A/A.sum(axis=1)[:,np.newaxis] 

假設

In [35]: A = np.random.rand(10,3) 
Out[35]: 
array([[ 0.26070074, 0.58940996, 0.78665012], 
     [ 0.7420538 , 0.72214655, 0.66633183], 
     [ 0.67189673, 0.67298124, 0.04628626], 
     [ 0.93935375, 0.45030544, 0.38292913], 
     [ 0.45410731, 0.26557299, 0.09573014], 
     [ 0.99872912, 0.31092656, 0.46294278], 
     [ 0.61108329, 0.71140089, 0.85548017], 
     [ 0.80012964, 0.64749927, 0.3292407 ], 
     [ 0.33229818, 0.01810878, 0.44460486], 
     [ 0.86525557, 0.0569463 , 0.43183502]]) 

你可以做以下的所有元素循環。

In [36]: A*(1/A.sum(axis=0)[np.newaxis,:] - 1/A.sum(axis=1)[:,np.newaxis]) 
Out[36]: 
array([[-0.12022572, -0.22751579, -0.3058817 ], 
     [-0.23713606, -0.17649948, -0.16474676], 
     [-0.3823249 , -0.33236228, -0.0229904 ], 
     [-0.38921906, -0.1527391 , -0.13097127], 
     [-0.48888156, -0.26594996, -0.09613741], 
     [-0.41381789, -0.10546217, -0.15833642], 
     [-0.18903571, -0.16660122, -0.20276794], 
     [-0.33044427, -0.21874513, -0.11216096], 
     [-0.36820095, -0.01870431, -0.46048657], 
     [-0.5094047 , -0.02924623, -0.22300407]])