Question

我希望使用以下公式更改数组中的所有值：

new_value = old_value * elec_space - elec_space

一个复杂的问题是阵列中48以上的所有值都会增加2，如49＆amp; 50将永远不会存在于原始数组中（infile，如下所示）。这意味着在执行上述计算之前，任何高于48的值都必须从中减去2。

原始值：

elec_space = 0.5

infile = 
[[41,   42,   43,   44]
 [41,   42,   44,   45]
 [41,   43,   45,   47]
 [44,   45,   46,   47]
 [44,   45,   47,   48]
 [44,   46,   48,   52]
 [47,   48,   51,   52]
 [47,   48,   52,   53]
 [47,   51,   53,   55]]

期望值：

infile =
[[ 20,  20.5,   21,  21.5]
 [ 20,  20.5,  21.5,  22]
 [ 20    21,    22,   23]
 [21.5,  22,   22.5,  23]
 [21.5   22,    23,  23.5]
 [21.5, 22.5,  23.5, 24.5]
 [ 23,  23.5,   24,  24.5]
 [ 23,  23.5,  24.5,  25]
 [ 23,   24,    25,   26]]

我试过了：

def remove_missing(infile):
    if infile > 48:
        return (infile - 2) * elec_space - elec_space
    else:
        return infile * elec_space - elec_space
A = remove_missing(infile[:,0])
B = remove_missing(infile[:,1])
M = remove_missing(infile[:,2])
N = remove_missing(infile[:,3])
infile = np.column_stack((A, B, M, N))

和

def remove_missing(infile):
    return (infile - 2) * elec_space - elec_space if infile > 50 else infile * elec_space - elec_space
A = remove_missing(infile[:,0])
B = remove_missing(infile[:,1])
M = remove_missing(infile[:,2])
N = remove_missing(infile[:,3])

但是每个人都获得了以下追溯：

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-181-dcc8e29a527f> in <module>()
      4     else:
      5         return infile * elec_space - elec_space
----> 6 A = remove_missing(infile[:,0])
      7 B = remove_missing(infile[:,1])
      8 M = remove_missing(infile[:,2])

<ipython-input-181-dcc8e29a527f> in remove_missing(infile)
      1 def remove_missing(infile):
----> 2     if infile > 48:
      3         return (infile - 2) * elec_space - elec_space
      4     else:
      5         return infile * elec_space - elec_space

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() 




---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-180-c407ec4fa95d> in <module>()
      2     return (infile - 2) * elec_space - elec_space if infile > 50 else infile * elec_space - elec_space
      3 
----> 4 A = remove_missing(infile[:,0])
      5 B = remove_missing(infile[:,1])
      6 M = remove_missing(infile[:,2])

<ipython-input-180-c407ec4fa95d> in remove_missing(infile)
      1 def remove_missing(infile):
----> 2     return (infile - 2) * elec_space - elec_space if infile > 50 else infile * elec_space - elec_space
      3 
      4 A = remove_missing(infile[:,0])
      5 B = remove_missing(infile[:,1])

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

我不认为a.any或a.all是正确的选项，因为我希望函数迭代运行数组列中的每一行，而不是根据其中一个更改所有值价值超过48。

有人有任何关于如何最好地解决这个问题的指示吗？

Answer 1

一种替代方法可能是从结果中减去48大于(infile - 2*(infile>48))* elec_space - elec_space的元素，如此 -

{{1}}

函数迭代地在数组的单个列中的每一行上工作 - numpy

1 个答案: