如何在Python中进行动态计算

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英文:

How to doing dynamic calculation in python

问题

如何使奇数索引执行(-)，偶数索引执行(+)，最大迭代次数为6。迭代1 +10，迭代2 -20，迭代3 +30，迭代4 -40，迭代5 +50，迭代6 -60

  1. import numpy as np
  3. AA = np.array([[9.27914],
  4.                [9.33246],
  5.                [9.26303],
  6.                [9.30597],
  7.                [9.6594],
  8.                [9.04283],
  9.                [8.88866],
  10.                [8.89956]])
  12. max_iter = 6
  14. for i in range(len(AA)):
  15.     if i % 2 == 0:
  16.         AA[i][0] = AA[i][0] + (i // 2 + 1) * 10
  17.     else:
  18.         AA[i][0] = AA[i][0] - (i // 2 + 1) * 10
  20. AA = np.round(AA, 5)  # 四舍五入到五位小数
  21. print("expected results:")
  22. print("AA=np.array([")
  23. for i in range(len(AA)):
  24.     print(f"    [{AA[i][0]:.5f}],")
  25. print("])")

预期结果：

  1. AA=np.array([
  2.     [19.27914],
  3.     [-10.66754],
  4.     [39.26303],
  5.     [-30.69403],
  6.     [59.6594],
  7.     [-50.95717],
  8.     [18.88866],
  9.     [-11.10044],
  10. ])

这段代码将按照您的要求对数组进行修改，使奇数索引执行减法，偶数索引执行加法，最大迭代次数为6，每次迭代加减10。希望这可以帮助您得到期望的结果。

英文:

How to make it like that so odd indexes will be doing (-) and even indexes will do (+) The max iteration is 6. iteration 1 +10, iteration 2 -20, iteration 3 +30, iteration 4 -40, iteration 5 + 50, iteration 6 -60

  1. AA = np.array([[9.27914]+10,
  2.               [9.33246]-20,
  3.               [9.26303]+30,
  4.               [9.30597]-40,
  5.               [9.6594 ]+50,
  6.               [9.04283]-60,
  7.               [8.88866]+10,
  8.               [8.89956]-20])

expected results:

  1. AA=np.array([
  2.     [19.27914],
  3.     [-10.66754],
  4.     [39.26303],
  5.     [-30.69403],
  6.     [59.6594],
  7.     [-50.95717],
  8.     [18.88866],
  9.     [-11.10044],
  10. ])

I try use this code but not working

  1. max_iter = 6
  2. iter = 0
  3. for i in range(len(AA)):
  4.     if i % 2 == 0:
  5.         AA[i][0] = AA[i][0] + (iter % max_iter)
  6.     else:
  7.        AA[i][0] = AA[i][0] - (iter % max_iter)
  8.     iter += 10

答案1

得分: 2

这是[tag:numpy]，不要循环，使用矢量化代码。

取模运算符（%）非常有用，可以重新启动一个序列。

输出：

  1. array([[ 19.27914],
  2.        [-10.66754],
  3.        [ 39.26303],
  4.        [-30.69403],
  5.        [ 59.6594 ],
  6.        [-50.95717],
  7.        [ 18.88866],
  8.        [-11.10044]])

使用的输入：

  1. AA = np.array([[9.27914],
  2.                [9.33246],
  3.                [9.26303],
  4.                [9.30597],
  5.                [9.6594 ],
  6.                [9.04283],
  7.                [8.88866],
  8.                [8.89956]])

中间步骤：

  1. (a%6+1)*10
  3. array([[10],
  4.        [20],
  5.        [30],
  6.        [40],
  7.        [50],
  8.        [60],
  9.        [10],
  10.        [20]])
  12. np.where(a%2, -1, 1)
  14. array([[ 1],
  15.        [-1],
  16.        [ 1],
  17.        [-1],
  18.        [ 1],
  19.        [-1],
  20.        [ 1],
  21.        [-1]])
英文:

This is [tag:numpy], don't loop, use vectorial code.

The modulo operator (%) is quite useful to restart a sequence.

  1. max_iter = 6
  3. a = np.arange(len(AA))[:,None]
  5. out = AA + (a%max_iter+1)*10 * np.where(a%2, -1, 1)

Output:

  1. array([[ 19.27914],
  2.        [-10.66754],
  3.        [ 39.26303],
  4.        [-30.69403],
  5.        [ 59.6594 ],
  6.        [-50.95717],
  7.        [ 18.88866],
  8.        [-11.10044]])

Used input:

  1. AA = np.array([[9.27914],
  2.                [9.33246],
  3.                [9.26303],
  4.                [9.30597],
  5.                [9.6594 ],
  6.                [9.04283],
  7.                [8.88866],
  8.                [8.89956]])

Intermediates:

  1. (a%6+1)*10
  3. array([[10],
  4.        [20],
  5.        [30],
  6.        [40],
  7.        [50],
  8.        [60],
  9.        [10],
  10.        [20]])
  12. np.where(a%2, -1, 1)
  14. array([[ 1],
  15.        [-1],
  16.        [ 1],
  17.        [-1],
  18.        [ 1],
  19.        [-1],
  20.        [ 1],
  21.        [-1]])

答案2

得分: 1

你已经非常接近了。只需进行3个小的更改。我在括号内添加了+1，对每个数组操作添加了*10，并将iter += 10更改为array += 1。

  1. max_iter = 6
  2. array = 0
  3. for i in range(len(AA)):
  4.     if i % 2 == 0:
  5.         AA[i][0] = AA[i][0] + (array % max_iter + 1) * 10
  6.     else:
  7.         AA[i][0] = AA[i][0] - (array % max_iter + 1) * 10
  8.     array += 1

实际上，你可以删除if else语句，如果使用以下代码可以在一行中完成：

  1. AA[i][0] = AA[i][0] + (-1) ** (i) * (array % max_iter + 1) * 10
英文:

You were very close. Just had to make 3 small changes. I added a +1 inside the parenthesis, added *10 for each of the array operations and change iter += 10 to array += 1

  1. max_iter = 6
  2. iter = 0
  3. for i in range(len(AA)):
  4.     if i % 2 == 0:
  5.         AA[i][0] = AA[i][0] + (iter % max_iter+1)*10
  6.     else:
  7.         AA[i][0] = AA[i][0] - (iter % max_iter+1)*10
  8.     iter += 1

In fact, you can remove the if else statement and do it in a single line if you use the following:

  1. AA[i][0] = AA[i][0] +(-1)**(i)* (iter % max_iter+1)*10

答案3

得分: 0

请尝试以下代码：

  1. AA = [
  2.     9.27914,
  3.     9.33246,
  4.     9.26303,
  5.     9.30597,
  6.     9.6594,
  7.     9.04283,
  8.     8.88866,
  9.     8.89956
  10. ]
  12. BB=[]
  14. iter = 10
  15. for i in range(len(AA)):
  16.     if i % 2 == 0:
  17.         var = AA[i] + (iter)
  18.     else:
  19.        var = AA[i] - (iter)
  20.     BB.append(var)
  21.     iter += 10
  22.     
  23. for i in range(len(BB)):
  24.     print(BB[i])

注意：最大迭代次数被省略了...将尽快修复。

英文:

Try this:

  1. AA = [
  2.     9.27914,
  3.     9.33246,
  4.     9.26303,
  5.     9.30597,
  6.     9.6594,
  7.     9.04283,
  8.     8.88866,
  9.     8.89956
  10. ]
  12. BB=[]
  14. iter = 10
  15. for i in range(len(AA)):
  16.     if i % 2 == 0:
  17.         var = AA[i] + (iter)
  18.     else:
  19.        var = AA[i] - (iter)
  20.     BB.append(var)
  21.     iter += 10
  22.     
  23. for i in range(len(BB)):
  24.     print(BB[i])

Note: left out max iterations... will fix asap.

答案4

得分: 0

你目前的根本问题是混淆了变量 iter 的含义。有时似乎你想要它是10的倍数，有时又是1到6之间的数字。让我们将其分离并稍微清理一下代码。

给定：

  1. import numpy
  2. AA = numpy.array([
  3.     [9.27914],
  4.     [9.33246],
  5.     [9.26303],
  6.     [9.30597],
  7.     [9.6594 ],
  8.     [9.04283],
  9.     [8.88866],
  10.     [8.89956]
  11. ])

然后我们可以这样做：

  1. max_iter = 6
  2. iter = 1
  3. for i in range(len(AA)):
  4.     delta = 10 * (iter % max_iter)
  5.     sign = 1 if i % 2 == 0 else -1
  6.     AA[i][0] += sign * delta
  7.     iter += 1
  9. print(AA)

以获得：

  1. [
  2.  [ 19.27914]
  3.  [-10.66754]
  4.  [ 39.26303]
  5.  [-30.69403]
  6.  [ 59.6594 ]
  7.  [  9.04283]
  8.  [ 18.88866]
  9.  [-11.10044]
  10. ]

或者如果你更喜欢接近你最初的代码：

  1. max_iter = 6
  2. iter = 1
  3. for i in range(len(AA)):
  4.     if i % 2 == 0:
  5.         AA[i][0] += 10 * (iter % max_iter)
  6.     else:
  7.         AA[i][0] -= 10 * (iter % max_iter)
  8.     iter += 1
  10. print(AA)
英文:

The root issue you have is a muddling of the meaning of your variable iter. Sometimes you seems to want to be a multiple of 10 and sometimes a number between 1 and 6. Let's separate that out and clean up the code a bit.

Given:

  1. import numpy
  2. AA = numpy.array([
  3.     [9.27914],
  4.     [9.33246],
  5.     [9.26303],
  6.     [9.30597],
  7.     [9.6594 ],
  8.     [9.04283],
  9.     [8.88866],
  10.     [8.89956]
  11. ])

Then we can do:

  1. max_iter = 6
  2. iter = 1
  3. for i in range(len(AA)):
  4.     delta = 10 * (iter % max_iter)
  5.     sign = 1 if i % 2 == 0 else -1
  6.     AA[i][0] += sign * delta
  7.     iter += 1
  9. print(AA)

to get:

  1. [
  2.  [ 19.27914]
  3.  [-10.66754]
  4.  [ 39.26303]
  5.  [-30.69403]
  6.  [ 59.6594 ]
  7.  [  9.04283]
  8.  [ 18.88866]
  9.  [-11.10044]
  10. ]

or if you like something closer to what you started with:

  1. max_iter = 6
  2. iter = 1
  3. for i in range(len(AA)):
  4.     if i % 2 == 0:
  5.         AA[i][0] += 10 * (iter % max_iter)
  6.     else:
  7.         AA[i][0] -= 10 * (iter % max_iter)
  8.     iter += 1
  10. print(AA)

答案5

得分: 0

以下是已翻译的代码部分：

  1. 我为你编写了一个不使用循环并能产生所需结果的函数。
  3.     AA = np.array([[9.27914],
  4.               [9.33246],
  5.               [9.26303],
  6.               [9.30597],
  7.               [9.6594 ],
  8.               [9.04283],
  9.               [8.88866],
  10.               [8.89956]])
  12.     expected_output = np.array([[19.27914],
  13.               [-10.66754],
  14.               [39.26303],
  15.               [-30.69403],
  16.               [59.6594],
  17.               [-50.95717],
  18.               [18.88866],
  19.               [-11.10044],
  20.     ])
  22.     def test(AA, iterstep, maxiter):
  23.         aa_shape = AA.shape
  24.         arr1 = np.tile(np.arange(1, max_iter+1), (aa_shape[0] // maxiter) + 1)[:aa_shape[0]] 
  25.         sign_arr = np.tile(np.array([1, -1]), (aa_shape[0] // 2) + 1)[:aa_shape[0]]
  26.         _ = (arr1*sign_arr)*10
  27.         _=_.reshape(AA.shape)
  28.         return AA + _
  30.     res = test(AA, 10, 6)
  32.     # 测试
  33.     assert (np.all(np.isclose(res, expected_output)))
英文:

I write for you a function that works without loops and give you desired result.

  1. AA = np.array([[9.27914],
  2.           [9.33246],
  3.           [9.26303],
  4.           [9.30597],
  5.           [9.6594 ],
  6.           [9.04283],
  7.           [8.88866],
  8.           [8.89956]])
  10. expected_output = np.array([[19.27914],
  11.           [-10.66754],
  12.           [39.26303],
  13.           [-30.69403],
  14.           [59.6594],
  15.           [-50.95717],
  16.           [18.88866],
  17.           [-11.10044],
  18. ])
  20. def test(AA, iterstep, maxiter):
  21.     aa_shape = AA.shape
  22.     arr1 = np.tile(np.arange(1, max_iter+1), (aa_shape[0] // maxiter) + 1)[:aa_shape[0]] 
  23.     sign_arr = np.tile(np.array([1, -1]), (aa_shape[0] // 2) + 1)[:aa_shape[0]]
  24.     _ = (arr1*sign_arr)*10
  25.     _=_.reshape(AA.shape)
  26.     return AA + _
  28. res = test(AA, 10, 6)
  30. # Test 
  31. assert (np.all(np.isclose(res, expected_output)))

huangapple
  • 本文由 发表于 2023年6月1日 22:21:59
  • 转载请务必保留本文链接：https://go.coder-hub.com/76382920.html
  • numpy
  • python
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