Python 在不使用 numpy 的情况下获得对角线？
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Get diagonal without using numpy?
提问by user3078618
I'm trying to get the diagonal from a matrix in Python without using numpy(I really can't use it). Does someone here knows how to do it?
我正在尝试从 Python 中的矩阵中获取对角线而不使用numpy（我真的无法使用它）。这里有人知道怎么做吗？
Example of what I want to get:
我想得到的例子：
get_diagonal ([[1,2,3,4],[5,6,7,8],[9,10,11,12]], 1, 1, 1)
Result: [1, 6, 11]
Or like:
或者像：
get_diagonal ([[1,2,3,4],[5,6,7,8],[9,10,11,12]], 1, 2, 1)
Result: [2, 7, 12]
Until know I've tried a lot of stuff but doesn't work.
直到知道我已经尝试了很多东西但没有用。
def obter_diagonal(matrix, line, column, direc):
d = []
if direc == 1:
for i in matrix:
for j in i:
if all(i == line, j == column):
d.extend(matrix[i][j])
else:
for i in matrix:
for j in i:
d.extend[len(matrix)-1-i][j]
return d
If direc==1I need to get the diagonal that goes from left-> right, top-> bottom.
If direc==-1need to get the diag that goes from right-> left, top->bottom.
如果direc==1我需要得到从左-> 右，上-> 下的对角线。
如果direc==-1需要获取从右-> 左、上-> 下的诊断。
采纳答案by jez
To get the leading diagonal you could do
要获得领先的对角线，您可以这样做
diag = [ mat[i][i] for i in range(len(mat)) ]
or even
甚至
diag = [ row[i] for i,row in enumerate(mat) ]
And play similar games for other diagonals. For example, for the counter-diagonal (top-right to bottom-left) you would do something like:
并为其他对角线玩类似的游戏。例如，对于反对角线（从右上角到左下角），您可以执行以下操作：
diag = [ row[-i-1] for i,row in enumerate(mat) ]
For other minor diagonals you would have to use ifconditionals in the list comprehension, e.g.:
对于其他小对角线，您必须if在列表理解中使用条件，例如：
diag = [ row[i+offset] for i,row in enumerate(mat) if 0 <= i+offset < len(row)]
回答by bcorso
def get_diagonal(m, i0, j0, d):
return [m[(i0 + i - 1)%len(m)][(j0 + d*i - 1)%len(m[0])]
for i in range(len(m))]
Which gets the diagonals in forward or reverse directions:
得到正向或反向对角线：
m = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]]
print get_diagonal(m, 1, 1, 1) # [1, 6, 11]
print get_diagonal(m, 1, 2, 1) # [2, 7, 12]
print get_diagonal(m, 1, 4,-1) # [4, 7, 10]
It even wraps around the matrix to get diagonals:
它甚至环绕矩阵以获得对角线：
print get_diagonal(m, 1, 4, 1) # [4, 5, 10]
print get_diagonal(m, 1, 1,-1) # [1, 8, 11]
print get_diagonal(m, 3, 1, 1) # [9, 2, 7 ]
回答by Shoaib Ali C H
Well, I have a solution that works for me.
好吧，我有一个适合我的解决方案。
Input :
输入 ：
First line contains an integer N
第一行包含一个整数 N
The next N lines denote the matrix's rows, with each line containing space-separated integers describing the columns.
接下来的 N 行表示矩阵的行，每行包含描述列的空格分隔的整数。
Sample Input :
样本输入：
3
11 2 4
4 5 6
10 8 -12
Code :
代码 ：
import sys
n = int(input().strip())
a = []
for a_i in range(n):
a_t = [int(a_temp) for a_temp in input().strip().split(' ')]
a.append(a_t)
pri_d = [];
pri_m = 0;
sec_d = [];
sec_m = n - 1;
for i in a:
pri_d.append(i[pri_m]);
sec_d.append(i[sec_m]);
pri_m = pri_m + 1;
sec_m = sec_m - 1;
print(pri_d);
print(sec_d);
output :
输出 ：
[11, 5, -12]
[4, 5, 10]
回答by Lean Bravo
Since nobody mentioned map or lambdas here, I'll leave a solution:
由于这里没有人提到 map 或 lambdas，我将留下一个解决方案：
list(map(lambda x: x[a.index(x)], a))
That way at array 0 it will grab element 0, and so on.
这样在数组 0 处，它将抓取元素 0，依此类推。
As for the opposite diagonal you might want to either flip the array bottom-up or take into consideration the length of the array minus one and subtract the current index to it:
至于相反的对角线，您可能想要自下而上翻转数组，或者考虑数组的长度减去一并减去当前索引：
list(map(lambda x: x[(len(a) - 1) - a.index(x)], a)))
Hope it helps!
希望能帮助到你！
回答by Rodrigo Carneiro
For diagonal:
对于对角线：
[m[i][i] for i in xrange(0, len(m))]
[m[i][i] for i in xrange(0, len(m))]
For counter-diagonal:
对于对角线：
[m[i][~i] for i in xrange(0, len(m))]
[m[i][~i] for i in xrange(0, len(m))]
