如何向NumPy数组中添加新数据

huangapple
go评论103阅读模式
英文:

How to add new data to numpy array

问题

Here's the translated code snippet without the translation of the code comments:

  1. 我想要将新数据添加到我的NumPy数组中。这是我的数组示例：
  3. A = np.array([
  4.     [4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
  5.     [5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
  6.     [5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
  7.     [4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
  8.     [4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
  9.     [4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
  10. ])
  12. B = np.array([
  13.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
  14. ])
  16. 如何在每2次迭代中将B添加到A中。
  18. 预期的A在添加B后如下所示：
  20. array([
  21.     [4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
  22.     [5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
  23.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  24.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  25.     [5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
  26.     [4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
  27.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  28.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  29.     [4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
  30.     [4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
  31.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  32.     [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  33. ])
  35. 我尝试使用了以下代码，但我希望它能够更动态。因为我想要使迭代动态化。现在我想每2次迭代，也许以后我想每50次迭代。这是我的代码：
  37. result = np.empty((0, A.shape[1]))
  39. for i in range(0, A.shape[0], 2):
  40.     result = np.vstack([result, A[i:i+2], B, B])
  42. print(result)

If you have any specific questions or need further assistance with this code, please let me know.

英文:

I want add new data to my numpy array. here is the example of my array.

  1. A = np.array([
  2. [4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
  3. [5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
  4. [5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
  5. [4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
  6. [4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
  7. [4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
  8. ])
  9. B = np.array([
  10. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
  11. ])

How to Add B each 2 iteration on A.

expected A after add B:

  1. array([
  2. [4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
  3. [5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
  4. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  5. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  6. [5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
  7. [4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
  8. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  9. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  10. [4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
  11. [4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
  12. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  13. [0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
  14. ])

I've try used this code but I want make it dynamic. because I want make the iteration dynamic. now I want each 2 iteration, maybe later I want make it each 50 iteration. here my code:

  1. result = np.empty((0, A.shape[1]))
  2. for i in range(0, A.shape[0], 2):
  3. result = np.vstack([result, A[i:i+2], B, B])
  4. print(result)

答案1

得分: 2

从较小的数组创建一个大的NumPy数组需要将所有数据从较小的数组复制过来，因此在循环中使用vstack构建新数组效率低下。相反，预先分配一个较大的数组，然后分配给切片：

  1. result = np.empty((A.shape[0] * 2, A.shape[1]))
  2. result[::4] = A[::2]
  3. result[1::4] = A[1::2]
  4. result[2::4] = B
  5. result[3::4] = B

现在，要软编码B行的频率和数量，您需要首先计算新数组的大小。由于我们每隔b_freq行（来自A）添加b_count行（来自B），可以计算new_row_count如下所示：

  1. b_freq = 3  # 每三行添加一次B
  2. b_count = 2 # 每b_freq行添加B 2次
  4. old_row_count = A.shape[0]
  5. new_row_count = old_row_count + int(np.ceil(old_row_count / b_freq) * b_count)
  7. step = b_freq + b_count
  9. result = np.empty((new_row_count, A.shape[1]))
  11. # 设置来自A的行
  12. for i in range(b_freq):
  13.     result[i::step] = A[i::step]
  15. # 设置来自B的行
  16. for i in range(b_count):
  17.     result[b_freq+i::step] = B

这将得到所需的结果。

另外，还可以创建一个包含result数组所有行的列表，并在完成列表后将行堆叠成数组：

  1. rows = []
  3. for i in range(0, A.shape[0], 2):
  4.     rows.extend([A[i:i+2], B, B])
  6. result = np.vstack(rows)

使用这种方法，软编码B行的频率更容易，因为不需要预先计算新数组的大小：只需更改循环中range的步长。

  1. b_freq = 3  # 每三行添加一次B
  2. b_count = 2 # 每b_freq行添加B 2次
  3. rows = []
  5. for i in range(0, A.shape[0], b_freq):
  6.     rows.append(A[i:i+b_freq])
  7.     rows.extend([B] * b_count)
  9. result = np.vstack(rows)

这将在每三行中产生两行零值。

英文:

Creating a big numpy array from smaller arrays requires you to copy all the data from the smaller arrays, so it's inefficient to keep building new arrays using vstack in a loop. Instead, pre-allocate a larger array, and then assign to the slices:

  1. result = np.empty((A.shape[0] * 2, A.shape[1]))
  2. result[::4] = A[::2]
  3. result[1::4] = A[1::2]
  4. result[2::4] = B
  5. result[3::4] = B

Now, to soft-code the frequency and number of B rows, you need to first calculate the size of the new array. Since we add b_count rows (from B) every b_freq rows of A, the new_row_count can be calculated as shown below:

  1. b_freq = 3 # Add B every three rows
  2. b_count = 2 # Add B 2 times every b_freq rows
  3. old_row_count = A.shape[0]
  4. new_row_count = old_row_count + int(np.ceil(old_row_count / b_freq) * b_count)
  5. step = b_freq + b_count
  6. result = np.empty((new_row_count, A.shape[1]))
  7. # Set rows from A
  8. for i in range(b_freq):
  9. result[i::step] = A[i::step]
  10. # Set rows from B
  11. for i in range(b_count):
  12. result[b_freq+i::step] = B

This gives the desired result:

  1. [[ 4.752735   1.16801   -3.9976401 -0.319121  -7.4322705 -3.8692   ]
  2. [ 5.2125     1.165454  -3.99143   -0.318649  -7.43615   -4.12229  ]
  3. [ 5.16795    1.083616  -3.9877    -0.315453  -7.43191   -3.97633  ]
  4. [ 0.         0.         0.         0.         0.         0.       ]
  5. [ 0.         0.         0.         0.         0.         0.       ]
  6. [ 4.711875   1.143996  -3.98545   -0.304904  -7.4152703 -3.90752  ]
  7. [ 5.2125     1.165454  -3.99143   -0.318649  -7.43615   -4.12229  ]
  8. [ 5.16795    1.083616  -3.9877    -0.315453  -7.43191   -3.97633  ]
  9. [ 0.         0.         0.         0.         0.         0.       ]
  10. [ 0.         0.         0.         0.         0.         0.       ]]

Alternatively, create a list containing all the rows of the result array, and stack the rows into an array after you've finished making the list:

  1. rows = []
  2. for i in range(0, A.shape[0], 2):
  3. rows.extend([A[i:i+2], B, B])
  4. result = np.vstack(rows)

With this method, soft-coding the frequency of B rows is easier since we don't need to pre-calculate the size of the new array: just change the step in the range of the for loop.

  1. b_freq = 3  # Add B every three rows
  2. b_count = 2 # Add B 2 times every b_freq rows
  3. rows = []
  4. for i in range(0, A.shape[0], b_freq):
  5. rows.append(A[i:i+b_freq])
  6. rows.extend([B] * b_count)
  7. result = np.vstack(rows)

Which gives two zero rows every three rows:

  1. [[ 4.752735    1.16801    -3.9976401  -0.319121   -7.4322705  -3.8692    ]
  2. [ 5.2125      1.165454   -3.99143    -0.318649   -7.43615    -4.12229   ]
  3. [ 5.16795     1.083616   -3.9877     -0.315453   -7.43191    -3.97633   ]
  4. [ 0.          0.          0.          0.          0.          0.        ]
  5. [ 0.          0.          0.          0.          0.          0.        ]
  6. [ 4.660315    0.89965    -3.98708    -0.321073   -7.41803    -3.69156   ]
  7. [ 4.839925    0.93606997 -3.9822102  -0.315018   -7.41289    -3.79918   ]
  8. [ 4.711875    1.143996   -3.98545    -0.304904   -7.4152703  -3.90752   ]
  9. [ 0.          0.          0.          0.          0.          0.        ]
  10. [ 0.          0.          0.          0.          0.          0.        ]]

huangapple
  • 本文由 发表于 2023年5月10日 23:51:44
  • 转载请务必保留本文链接：https://go.coder-hub.com/76220446.html
  • numpy
  • python
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定