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[英]Replace the diagonal elements of a matrix with sum of other elements in the row in Python
[英]Sum of diagonal elements in a matrix
我试图找出矩阵中对角线元素的总和。 这里,n 是方阵的大小,a 是矩阵。 有人可以向我解释这里发生了什么。
n = 3
a = [[11,2,4],[4,5,6],[10,8,-12]]
sum_first_diagonal = sum(a[i][i] for i in range(n))
sum_second_diagonal = sum(a[n-i-1][n-i-1] for i in range(n))
print(str(sum_first_diagonal)+" "+str(sum_first_diagonal))
使用对任何矩阵计算都非常强大的 numpy 库。 对于您的具体情况:
import numpy as np
a = [[11,2,4],[4,5,6],[10,8,-12]]
b = np.asarray(a)
print('Diagonal (sum): ', np.trace(b))
print('Diagonal (elements): ', np.diagonal(b))
您可以使用 pip 或您在许多网站上找到的其他方式轻松安装 numpy。
如果您想要所有对角线,而不仅仅是主对角线,请检查这个也使用 numpy.
编辑
mhawke,如果要计算对角线(次对角线),如wikipedia中所述,您可以在 numpy 中翻转矩阵
import numpy as np
a = [[11,2,4],[4,5,6],[10,8,-12]]
b = np.asarray(a)
b = np.fliplr(b)
print('Antidiagonal (sum): ', np.trace(b))
print('Antidiagonal (elements): ', np.diagonal(b))
试试这个求和你的第二个对角线:
sum(a[i][n-i-1] for i in range(n))
内部循环访问这些条目:
>>> n = 3
>>> [(i, n-i-1) for i in range(n)]
[(0, 2), (1, 1), (2, 0)]
您的样本矩阵的这条对角线的总和值为:
>>> n = 3
>>> sum(a[i][n-i-1] for i in range(n))
19
您的代码中的错误是对两个维度使用相同的表达式:
a[n-i-1][n-i-1]
它将以相反的顺序再次处理第一个对角线[(2, 2), (1, 1), (0, 0)]
给你两次相同的总和。
从方阵中获取总和和对角线和
squared_matrix = [[2,3,4],[4,3,3],[3,3,4]]
s, ds = get_sum(squared_matrix)
def get_sum(diag_mat):
n = len(diag_mat)
total = sum([diag_mat[i][j] for i in range(n) for j in range(j)]
d_sum = sum([diag_mat[i][j] if i==j else 0 for i in range(n) for j in range(j)]
return d_sum, total
尝试这个:
n=3
sum_second_diagonal=sum([a[i][j] for i in range(n) for j in range(n) if i==j]) #it will add when i==j
您可以按照以下代码获取矩阵中第一对角线和第二对角线之和的总和。 我认为这将是最明白的方式。
得到第一对角线的总和,
n = 3
a = [[11,2,4],[4,5,6],[10,8,-12]]
sum_1_diagonal = 0
for x in range(0, n):
for y in range(0, n):
if x == y:
sum_1_diagonal = sum_1_diagonal + a[x][y]
得到第二对角线的总和,
n = 3
a = [[11,2,4],[4,5,6],[10,8,-12]]
sum_2_diagonal = 0
for x in range(0, n):
for y in range(0, n):
if x + y == n - 1:
sum_2_diagonal = sum_2_diagonal + a[x][y]
def sum_up_diagonals(li):
index = len(li)
first_dia = sum(li[i][i]for i in range(index))
second_dia = sum(li[i][index-i-1]for i in range(index))
return (first_dia,second_dia)
传递你的清单。 这应该适合你:)
O(n) 时间解来找到给定多维数组的对角线差。
def diagonalDifference(arr):
# arr[0][0], arr[1][1], arr[2][2]
# arr[0][2], arr[1][1], arr[2][0]
sumOfDiagonalFromLeft = 0
sumOfDiagonalFromRight = 0
pointIndexFromLeft = 0
pointIndexFromLast = len(arr)-1
for i in range(len(arr)):
sumOfDiagonalFromLeft += arr[i][pointIndexFromLeft]
# print(arr[i][pointIndexFromLeft])
pointIndexFromLeft += 1
for i in range(len(arr)):
sumOfDiagonalFromRight += arr[i][pointIndexFromLast]
# print(arr[i][pointIndexFromLast])
if pointIndexFromLast < 0:
break
else:
pointIndexFromLast -= 1
diagonalDifference = abs(sumOfDiagonalFromLeft - sumOfDiagonalFromRight)
return diagonalDifference
arr = [[11, 2, 4], [4, 5, 6], [10, 8, -12]]
print(diagonalDifference(arr))
//we simply swap corner elements of matrix
def diagonalDifference(arr):
c=0;d=0
for i in range(len(arr)):
for j in range(len(arr)):
if i==j :
c = c+ arr[i][j]
for k in range(len(arr)):
arr[i][k-1],arr[i][k-2] = arr[i][k-2],arr[i][k-1]
for j in range(len(arr)):
if i==j :
d = d + arr[i][j]
for k in range(len(arr)):
arr[i][k-1],arr[i][k-2] = arr[i][k-2],arr[i][k-1]
print(c,d)
result = abs(c-d)
print(result)
将矩阵的动态输入作为列表输入的 Python 3 解决方案
#Even matrix size is dynamic in this code as "n".
n=int(input())
s1=0
s2=0
a =[]
for i in range(n):
# x here take input of size n and as separate lists to act like a matrix.
x=list(map(int,input().strip().split()))[:n]
a.append(x)
for i in range(n):
s1+=a[i][i]
s2+=a[i][n-i-1]
# If use abs() only if non-negative output is needed!!
print(abs(s1-s2))
# First enter the size of matrix then enter the matrix vales with spaces in each line
假设一个方阵 (nxn),您可以计算主对角线和次对角线的总和,只需对矩阵的行进行 1 次迭代; 通过跟踪每次计算中涉及的索引。
mat = [
[1,2,3],
[4,5,6],
[9,8,9]
]
primary = 0
secondary = 0
low = 0 # start index for primary
high = len(mat)-1 # end index for secondary
for row in mat:
primary += row[low]
secondary += row[high]
low += 1
high -= 1
print(f'Primary Sum = {primary}', f'Secondary Sum = {secondary}', sep='\n')
假设,您有一个矩阵 A n×m。
A = [[random.rand() for i in range(n+1)] for i in range(m+1)]
d = m+n-1 # that's the overall number of different diagonals
sumrow = [0]*d # I'd like to create a list of all the sums of those
for diag in range(d): # iterate by diagonal number
# A difference of two related spike functions, so that it was a spike without the top
diaglength = int(max(0,1+d/2-abs(diag+1/2-d/2)) - max(0,1+d/2-n-abs(diag+1/2-d/2)))
# initial coordinates for further iteration inside sum() function
x = max(0, diag-m+1)
y = max(0, m-diag-1)
# iterating values along the current diagonal
sumrow[diag]=sum(A[x+i][y+i] for i in range(0,diaglength))
因此, sumrow[] 是每个对角线的每个总和的列表,请选择不要弄乱。 我认为,很明显要切断这段代码以立即获得一些特定的价值。 从技术上讲,这段代码可以压缩成一行。 真的很长的线。
但为什么?
mat = [[1 ,2, 3, 4,5],[1,2,1,2,3],[2,3,4,5,2],[1,3,5,1,1] ,[1,2,1,2,1]]
def isSquare(mat): if len(mat) == 1: return True for i in range(len(mat)): if len(mat[i]) != len(mat): print("Matrix is not Squared ") 返回 False 返回 True
def sumOfLeftDiagonal(mat): d1 = 0 size = len(mat) for i in range(0,size): for j in range(0,size): if i == j: d1 = d1 + mat[i][ j] 返回 d1
def sumOfRightDiagonal(mat): d2 = 0 size = len(mat) for i in range(0,size): d2 = d2 + mat[i][(size-1)-i] return d2
if isSquare(mat): if len(mat) == 1: print("Difference: ",mat[0]) else: print("sumOfLeftDiagonal: ",sumOfLeftDiagonal(mat)) print("sumOfRightDiagonal: ",sumOfRightDiagonal( mat)) print("差异:",abs(sumOfLeftDiagonal(mat)-sumOfRightDiagonal(mat)))
这是获取完整primary diagonal
和secondary diagonal
的更简单方法
squared_matrix = [
[2, 3, 4],
[4, 3, 3],
[3, 3, 4]
]
def primary_diagonal(matrix):
sum = 0
for i in range(len(matrix)):
sum += matrix[i][i]
return sum
def secondary_left_diagonal(matrix):
sum = 0
for i in range(len(matrix)):
sum += matrix[i][len(matrix) - i - 1]
return sum
print(primary_diagonal(squared_matrix))
print(secondary_left_diagonal(squared_matrix))
结果
9
10
这是一种方法:
由于矩阵是正方形的,我们可以通过使用原始列表和反向列表来获得对角线和反对角线的总和。
matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
to_sum =[] # an empty list for collection values
for i in range(len(matrix)): # diagonal coordinats [0][0], [1][1].... [n][n]
to_sum.append(matrix[i][i]) # append value of diagonal
rev = matrix[i][::-1] # for anti-diagonal reverse list
to_sum.append(rev[i]) # append value of anti-diagonal
# to_sum list [1, 4, 6, 7, 11, 10, 16, 13]
return sum(to_sum) # return sum of all values in list (68)
由于您知道第i
行的对角线元素的位置,因此您可以将其写得很密集,如下所示:
d = sum(row[i] + row[-1-i] for i, row in a)
而且,对于奇数大小的矩阵,您不应该两次添加中心元素:
if len(a)%2:
centre = len(a)//2
d -= a[centre][centre]
def sum_diagnol():
import random
sum=0
r1=int(input("row"))
c1=int(input("col"))
a=[[random.random()for col in range(c1)]for row in range(r1)]
print("enter elements")
for i in range(r1):
for j in range(c1):
a[i][j]=int(input("enter elements"))
r2=int(input("row"))
c2=int(input("col"))
b=[[random.random()for col in range(c2)]for row in range(r2)]
print("enter elements")
for i in range(r2):
for j in range(c2):
b[i][j]=int(input("enter elements"))
c=[[random.random()for col in range(c2)]for row in range(r1)]
if(c1==r2):
for i in range(r1):
for j in range(c2):
c[i][j]=0
for k in range(c2):
c[i][j]=a[j][k]*b[k][j]
else:
print("multiplication not possible")
for i in range(r1):
for j in range(c2):
print(c[i][j],end=" ")
print()
sum_diagnol()
我找到了一个简单的算法来完成这项任务。
将方阵存储在单个一维数组中。
让'n'是方阵的阶。
有两条对角线,一条从顶行最左边的元素开始,另一个从顶行的第 n 个元素开始。
从包含矩阵中所有数字的数组中获取从顶行最左边的元素开始的对角线上数字的索引; 只需从索引 1 开始递归地添加 (n+1)。也就是说,数组中从左到右对角线的元素的索引为 1, 1+(n+1) , (n+2)+(n+1) , (2n+3)+(n+1) 直到数组的最后一个索引。
从包含矩阵中所有数字的数组中获取另一个对角线数字的索引; 只需将 (n-1) 递归添加到从 index 开始的索引等于“n”,这是方阵的顺序。 也就是说,数组中从右到左对角线元素的索引是,n,n+(n-1),(2n-1)+(n-1) 等等,直到索引等于'数组的长度 - (n-1)'。
如果顺序是奇数,则从最终总和中减去数组中的中间数。
示例“c++”代码如下:
#include<iostream>
using namespace std;
int sumOfDiagonalNumbersInSquareMatrix(int numberArray[],int order){
int sumOfLeftToRightDiagonal = 0;
int sumOfRightToLeftDiagonal = 0;
int length = order*order;
for(int i=0; i<length;i+=(order+1)){
//cout<<numberArray[i]<<"\t";
sumOfLeftToRightDiagonal = sumOfLeftToRightDiagonal + numberArray[i];
}
for(int i=(order-1);i<=length-order;i+=(order-1)){
//cout<<numberArray[i]<<"\t";
sumOfRightToLeftDiagonal = sumOfRightToLeftDiagonal + numberArray[i];
}
if(order % 2 != 0){
return (sumOfLeftToRightDiagonal + sumOfRightToLeftDiagonal) - numberArray[(length/2)];
}
return (sumOfLeftToRightDiagonal + sumOfRightToLeftDiagonal);
}
int main(){
int nums[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
cout<<sumOfDiagonalNumbersInSquareMatrix(nums,4);
return 0;
}
你可以在这里运行它:http: //cpp.sh/6cmdp
我不明白为什么没有人发布任何好的解决方案。 这是下降解决方案:
length = len(arr)
r1 = 0
r2 = 0
for i in range(length):
r1 += arr[i][length - i - 1]
r2 += arr[i][i]
print(r1 + r2)
# If you want sum of there absolute values
print(abs(r1) + abs(r2))
这里 arr 是一个二维列表。
'''
a = [[],[],[]] #your matrix
s = 0
for i in range(len(a)):
for j in range(len(a[0])):
if i == j:
s += a[i][j]
print('sum ='s)
''' 这里是一个简单的方法。 谢谢
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