如何找到Python中矩阵最大数量的索引?

时间:2019-03-17 03:05:34

标签: python python-3.x

m ->我的矩阵

m = [[19, 17, 12], [6, 9, 3], [8, 11, 1], [18, 1, 12]]

最大->我已经找到了最大的数字

max = 19 

现在我找不到索引

for i in range(len(m)):
  for c in m[i]:
    if c==19:
       print(m.index(c))

我遇到了错误

Traceback (most recent call last):
  File "<pyshell#97>", line 4, in <module>
    print(m.index(c))
ValueError: 19 is not in list

我该如何处理?

4 个答案:

答案 0 :(得分:2)

从我的个人“备忘单”中,或根据“ HS-星云”的建议numpy docs

import numpy as np

mat = np.array([[1.3,3.4,0.1],[4.0,3.2,4.5]])

i, j = np.unravel_index(mat.argmax(), mat.shape)
print(mat[i][j])

# or the equivalent:
idx = np.unravel_index(mat.argmax(), mat.shape)
print(mat[idx])

答案 1 :(得分:0)

您需要使用numpy。这是一个工作代码。使用numpy.array,您可以从中进行许多计算。

import numpy as np
mar = np.array([[19, 17, 12], [6, 9, 3], [8, 11, 1], [18, 1, 12]])
# also OK with
# mar = [[19, 17, 12], [6, 9, 3], [8, 11, 1], [18, 1, 12]]
test_num = 19   # max(mar.flatten()) --> 19
for irow, row in enumerate(mar):
    #print(irow, row)
    for icol, col in enumerate(row):
        #print(icol, col)
        if col==test_num:
            print("** Index of {}(row,col): ".format(test_num), irow, icol)

输出将是:

** Index of 19(row,col):  0 0

如果您使用test_num = 11,则会得到** Index of 11(row,col): 2 1

答案 2 :(得分:0)

您不需要numpy,就可以同时搜索max和索引。

m = [[19, 17, 12], [6, 9, 3], [8, 11, 1], [18, 1, 12]]
max_index_row = 0
max_index_col = 0
for i in range(len(m)):
  for ii in range(len(m[i])):
    if m[i][ii] > m[max_index_row][max_index_col]:
      max_index_row = i
      max_index_col = ii
print('max at '+str(max_index_row)+','+str(max_index_col)+'('+str(m[max_index_row][max_index_col])+')')

输出: max at 0,0(19)

m = [[19, 17, 12], [20, 9, 3], [8, 11, 1], [18, 1, 12]]

max at 1,0(20)

答案 3 :(得分:0)

使用numpy更简单。您可以使用以下代码在矩阵(数组)中找到最大值的坐标(xi,yi):

import numpy as np
m = np.array([[19, 17, 12], [6, 9, 3], [8, 11, 1], [18, 1, 12]])
i = np.unravel_index(np.argmax(m), m.shape)