index_array = index_array([1:index_temp-1 index_temp+1:end]);
时间: 2024-05-28 08:13:01 浏览: 9
这行代码是在MATLAB中进行数组操作,用于删除数组中指定索引位置的元素。具体解释如下:
- `index_array`:要进行操作的数组。
- `[1:index_temp-1 index_temp+1:end]`:创建一个新的索引数组,其中不包括原数组中索引位置为`index_temp`的元素。`1:index_temp-1`表示从第一个元素到`index_temp-1`位置的元素,`index_temp+1:end`表示从`index_temp+1`位置到最后一个元素的元素。
- `index_array([1:index_temp-1 index_temp+1:end])`:使用新的索引数组来选择原数组中需要保留的元素,从而删除了索引位置为`index_temp`的元素。
相关问题
vector<BossFacesDirData> boss_face_dir_data_vec; boss_face_dir_data_vec.clear(); for (int a = 0; a < boss_faces.size(); a++) { boss_face_dir_data.dir_num = 0; double temp_dir1[3]; getFaceDir(boss_faces[a], temp_dir1); int is_eq = 0; for (int b = 0; b < boss_faces_dirs.size(); b++) { double temp_dir2[3]; temp_dir2[0] = boss_faces_dirs[b].X; temp_dir2[1] = boss_faces_dirs[b].Y; temp_dir2[2] = boss_faces_dirs[b].Z; UF_VEC3_is_equal(temp_dir1, temp_dir2, 0.001, &is_eq); if (is_eq) { boss_face_dir_data.dir_num = b + 1; } } if (boss_face_dir_data.dir_num > 20) { break; } UF_VEC3_copy(temp_dir1, boss_face_dir_data.face_dir); boss_face_dir_data_vec.push_back(boss_face_dir_data); continue; 改进这段代码
根据你提供的代码,我对其进行了一些改进。主要的改进包括:
1. 使用 `std::vector` 的 `emplace_back` 方法来避免不必要的复制操作。
2. 使用 `std::find_if` 算法来查找具有相同方向的面的索引。
3. 使用 `std::unordered_map` 来存储面的方向向量,以便进行快速查找。
下面是改进后的代码:
```cpp
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
struct BossFacesDirData {
int dir_num;
double face_dir[3];
};
bool isEqual(const double* dir1, const double* dir2) {
return (std::abs(dir1[0] - dir2[0]) < 0.001 &&
std::abs(dir1[1] - dir2[1]) < 0.001 &&
std::abs(dir1[2] - dir2[2]) < 0.001);
}
int main() {
std::vector<BossFacesDirData> boss_face_dir_data_vec;
boss_face_dir_data_vec.reserve(boss_faces.size());
std::unordered_map<std::array<double, 3>, int> boss_faces_dirs_map;
for (int b = 0; b < boss_faces_dirs.size(); b++) {
boss_faces_dirs_map[{boss_faces_dirs[b].X, boss_faces_dirs[b].Y, boss_faces_dirs[b].Z}] = b + 1;
}
for (int a = 0; a < boss_faces.size(); a++) {
double temp_dir1[3];
getFaceDir(boss_faces[a], temp_dir1);
auto it = std::find_if(boss_faces_dirs_map.begin(), boss_faces_dirs_map.end(),
[&](const auto& pair) { return isEqual(pair.first.data(), temp_dir1); });
BossFacesDirData boss_face_dir_data;
boss_face_dir_data.dir_num = (it != boss_faces_dirs_map.end()) ? it->second : 0;
std::copy_n(temp_dir1, 3, boss_face_dir_data.face_dir);
boss_face_dir_data_vec.emplace_back(std::move(boss_face_dir_data));
if (boss_face_dir_data.dir_num > 20) {
break;
}
}
// 输出分组结果
for (const auto& data : boss_face_dir_data_vec) {
std::cout << "Dir Num: " << data.dir_num << ", Face Dir: " << data.face_dir[0] << ", "
<< data.face_dir[1] << ", " << data.face_dir[2] << std::endl;
}
return 0;
}
```
请注意,上述代码中的 `isEqual` 函数用于比较两个方向向量是否相等。你可能需要根据你的具体需求进行适当的修改。此外,我假设你已经实现了 `getFaceDir` 函数来获取面的方向向量,并且 `boss_faces` 和 `boss_faces_dirs` 分别表示面和面方向的数据。
clc,clear; load('b.mat'); Q=vp(10); pp=vp(2); global space space = zeros(1e7,12); global ans_num ans_num = 0; global DB DB = vp(12); for v1 = 1:8 for v2 = v1+1:9 L1 = v2 - v1 - 1; L2 = 12 - v2; L = L1 + 2*L2; al = 2^L; for i=1:al aa = de2bi(i-1,L,2,'left-msb'); E = zeros(2,12); E(1,v1) = 1; E(2,v2) = 1; for j=1:L1 E(1,v1+j) = aa(j); end for j=1:L2 E(1,v2+j) = aa(L1+j); E(2,v2+j) = aa(L1+L2+j); end pe = E(1,:) | E(2,:); penum = 0; for j=v2+1:12 penum = penum + pe(j); end if(penum > 9-v2) continue; end re = mod(pp * E(:,:),2); w = [v1 v2]; s = zeros(1,12); d = [1 2 3 4 5 6 7 8 9 10 11 12]; d(w) = []; s(v1) = 0; s(v2) = 0; t0 = 0; t1 = 0; T0 = zeros(1024,12); T1 = zeros(1024,12); for t = 1:1024 s(d) = Q(t,:); h=[]; for ii=1:4 rre = 0; for jj=1:12 rre = rre * 2 + mod(s(jj)+re(ii,jj),2); end h = [h rre]; end if(b(h(1)+1) == b(h(2)+1) && b(h(2)+1) == b(h(3)+1) && b(h(3)+1) == b(h(4)+1)) if(b(h(1)+1) == 0) t0 = t0+1; T0(t0,:) = s(:); else t1 = t1+1; T1(t1,:) = s(:); end end end combine_2(T0,T1,E,2,t0,t1); end end end
这是一段MATLAB代码,我可以将其转换为Python代码。以下是转换后的代码:
```python
import numpy as np
from numpy.matlib import repmat
from scipy.io import loadmat
b_data = loadmat('b.mat')
vp_data = loadmat('vp.mat')
b = b_data['b'][0]
Q = vp_data['vp'][9]
pp = vp_data['vp'][1]
DB = vp_data['vp'][11]
space = np.zeros((1e7, 12))
ans_num = 0
def de2bi(num, n, order, msb):
if order == 2:
b_str = bin(num)[2:].zfill(n)
else:
b_str = np.base_repr(num, base=order, padding=n)
if msb == 'left-msb':
b = np.array([int(b_str[i]) for i in range(n)])
else:
b = np.array([int(b_str[n - i - 1]) for i in range(n)])
return b
def combine_2(T0, T1, E, num, t0, t1):
global space
for i in range(1, t0 + 1):
for j in range(1, t1 + 1):
if np.all(T0[i, :] == T1[j, :]):
continue
temp = T0[i, :] + T1[j, :]
if np.any(temp > 1):
continue
temp = np.append(temp, np.zeros((1,)))
if np.any(temp[E == 1] == 0):
continue
temp = temp[:12]
if np.any(temp > 1):
continue
if np.any(temp[E == 1] == 0):
continue
global ans_num
ans_num += 1
space[ans_num, :] = temp
if ans_num % 100000 == 0:
print(ans_num)
for v1 in range(1, 9):
for v2 in range(v1 + 1, 10):
L1 = v2 - v1 - 1
L2 = 12 - v2
L = L1 + 2 * L2
al = 2 ** L
for i in range(al):
aa = de2bi(i - 1, L, 2, 'left-msb')
E = np.zeros((2, 12))
E[0, v1] = 1
E[1, v2] = 1
for j in range(L1):
E[0, v1 + j + 1] = aa[j]
for j in range(L2):
E[0, v2 + j + 1] = aa[L1 + j]
E[1, v2 + j + 1] = aa[L1 + L2 + j]
pe = E[0, :] | E[1, :]
penum = pe[v2:].sum()
if penum > 9 - v2:
continue
re = pp * repmat(E, 2, 1)
w = np.array([v1, v2])
s = np.zeros((1, 12))
d = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
d[w - 1] = 0
d = d[d != 0]
s[0, v1 - 1] = 0
s[0, v2 - 1] = 0
t0 = 0
t1 = 0
T0 = np.zeros((1024, 12))
T1 = np.zeros((1024, 12))
for t in range(1, 1025):
s[0, d - 1] = Q[t - 1, :]
h = np.array([])
for ii in range(4):
rre = 0
for jj in range(12):
rre = rre * 2 + np.mod(s[0, jj] + re[ii, jj], 2)
h = np.append(h, rre)
if b[int(h[0])] == b[int(h[1])] == b[int(h[2])] == b[int(h[3])]:
if b[int(h[0])] == 0:
t0 += 1
T0[t0, :] = s
else:
t1 += 1
T1[t1, :] = s
combine_2(T0, T1, E, 2, t0, t1)
```
请注意,由于Python的语法和MATLAB略有不同,因此在转换过程中可能需要进行一些修改。
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