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:
我想要将新数据添加到我的NumPy数组中。这是我的数组示例：
A = np.array([
[4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
[5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
[5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
[4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
[4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
[4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
])
B = np.array([
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
])
如何在每2次迭代中将B添加到A中。
预期的A在添加B后如下所示：
array([
[4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
[5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
[4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
[4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
])
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：
result = np.empty((0, A.shape[1]))
for i in range(0, A.shape[0], 2):
result = np.vstack([result, A[i:i+2], B, B])
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.
A = np.array([
[4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
[5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
[5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
[4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
[4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
[4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
])
B = np.array([
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
])
How to Add B each 2 iteration on A.
expected A after add B:
array([
[4.752735, 1.16801, -3.9976401, -0.319121, -7.4322705, -3.8692],
[5.2125, 1.165454, -3.99143, -0.318649, -7.43615, -4.12229],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[5.16795, 1.083616, -3.9877, -0.315453, -7.43191, -3.97633],
[4.660315, 0.89965, -3.98708, -0.321073, -7.41803, -3.69156],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[4.839925, 0.93606997, -3.9822102, -0.315018, -7.41289, -3.79918],
[4.711875, 1.143996, -3.98545, -0.304904, -7.4152703, -3.90752]
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
])
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:
result = np.empty((0, A.shape[1]))
for i in range(0, A.shape[0], 2):
result = np.vstack([result, A[i:i+2], B, B])
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：
result = np.empty((A.shape[0] * 2, A.shape[1]))
result[::4] = A[::2]
result[1::4] = A[1::2]
result[2::4] = B
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：
b_freq = 3 # 每三行添加一次B
b_count = 2 # 每b_freq行添加B 2次
old_row_count = A.shape[0]
new_row_count = old_row_count + int(np.ceil(old_row_count / b_freq) * b_count)
step = b_freq + b_count
result = np.empty((new_row_count, A.shape[1]))
# 设置来自A的行
for i in range(b_freq):
result[i::step] = A[i::step]
# 设置来自B的行
for i in range(b_count):
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：
rows = []
for i in range(0, A.shape[0], 2):
rows.extend([A[i:i+2], B, B])
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。
b_freq = 3 # 每三行添加一次B
b_count = 2 # 每b_freq行添加B 2次
rows = []
for i in range(0, A.shape[0], b_freq):
rows.append(A[i:i+b_freq])
rows.extend([B] * b_count)
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:
result = np.empty((A.shape[0] * 2, A.shape[1]))
result[::4] = A[::2]
result[1::4] = A[1::2]
result[2::4] = B
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:
b_freq = 3 # Add B every three rows
b_count = 2 # Add B 2 times every b_freq rows
old_row_count = A.shape[0]
new_row_count = old_row_count + int(np.ceil(old_row_count / b_freq) * b_count)
step = b_freq + b_count
result = np.empty((new_row_count, A.shape[1]))
# Set rows from A
for i in range(b_freq):
result[i::step] = A[i::step]
# Set rows from B
for i in range(b_count):
result[b_freq+i::step] = B
This gives the desired result:
[[ 4.752735 1.16801 -3.9976401 -0.319121 -7.4322705 -3.8692 ]
[ 5.2125 1.165454 -3.99143 -0.318649 -7.43615 -4.12229 ]
[ 5.16795 1.083616 -3.9877 -0.315453 -7.43191 -3.97633 ]
[ 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. ]
[ 4.711875 1.143996 -3.98545 -0.304904 -7.4152703 -3.90752 ]
[ 5.2125 1.165454 -3.99143 -0.318649 -7.43615 -4.12229 ]
[ 5.16795 1.083616 -3.9877 -0.315453 -7.43191 -3.97633 ]
[ 0. 0. 0. 0. 0. 0. ]
[ 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:
rows = []
for i in range(0, A.shape[0], 2):
rows.extend([A[i:i+2], B, B])
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.
b_freq = 3 # Add B every three rows
b_count = 2 # Add B 2 times every b_freq rows
rows = []
for i in range(0, A.shape[0], b_freq):
rows.append(A[i:i+b_freq])
rows.extend([B] * b_count)
result = np.vstack(rows)
Which gives two zero rows every three rows:
[[ 4.752735 1.16801 -3.9976401 -0.319121 -7.4322705 -3.8692 ]
[ 5.2125 1.165454 -3.99143 -0.318649 -7.43615 -4.12229 ]
[ 5.16795 1.083616 -3.9877 -0.315453 -7.43191 -3.97633 ]
[ 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. ]
[ 4.660315 0.89965 -3.98708 -0.321073 -7.41803 -3.69156 ]
[ 4.839925 0.93606997 -3.9822102 -0.315018 -7.41289 -3.79918 ]
[ 4.711875 1.143996 -3.98545 -0.304904 -7.4152703 -3.90752 ]
[ 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. ]]
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