Python Expected 1D or 2D array, got 3D array instead 関数作成でのエラー

Question

前提・実現したいこと

関数を初めて作ってい見たのですが
いまいちよく分かりません。
下記のコードの場合は、何がいけないのでしょうか？

該当のソースコード

#===========================================================================================
theta = math.radians(0)
number_probe_xy = 21
width = 1
pitch_xy = (float(width)/(number_probe_xy-1))
############################################################################################


############################################################################################
#===========================================================================================
#
#                                  　　　　座標計算の関数定義
#
#===========================================================================================
def prob_locat (z):
    lst=[]
    for i in range(number_probe_xy):
    #上面(+y側の面)
        x = dec.Decimal((0.5 - pitch_xy*(i))*np.cos(theta))  -  dec.Decimal((0.5)*np.sin(theta))
        y = dec.Decimal((0.5 - pitch_xy*(i))*np.sin(theta))  +  dec.Decimal((0.5)*np.cos(theta))
        z = z
        lst += [x, y, z]
    #風上面   
    for i in range(number_probe_xy):
        x = dec.Decimal((-0.5)*np.cos(theta))  -  dec.Decimal((0.5 - pitch_xy*(i))*np.sin(theta))
        y = dec.Decimal((-0.5)*np.sin(theta))  +  dec.Decimal((0.5 - pitch_xy*(i))*np.cos(theta))
        z = z
        lst += [x, y, z]
    #下面(-y側の面)
    for i in range(number_probe_xy):
        x = dec.Decimal((-0.5 + pitch_xy*(i))*np.cos(theta))  -  dec.Decimal((-0.5)*np.sin(theta))
        y = dec.Decimal((-0.5 + pitch_xy*(i))*np.sin(theta))  +  dec.Decimal((-0.5)*np.cos(theta))
        z = z
        lst += [x, y, z]
    #風下面
    for i in range(number_probe_xy):
        x = dec.Decimal((0.5)*np.cos(theta))  -  dec.Decimal((-0.5 + pitch_xy*(i))*np.sin(theta))
        y = dec.Decimal((0.5)*np.sin(theta))  +  dec.Decimal((-0.5 + pitch_xy*(i))*np.cos(theta))
        z = z
        lst += [x, y, z]

    ##上記のリストを3列にする関数
    def convert_1d_to_2d(lst, cols):
        return [lst[i:i + cols] for i in range(0, len(lst), cols)]
    lst_reshape = convert_1d_to_2d(lst, 3)
    return lst_reshape



test = []
test.append(prob_locat(0.5))
np.savetxt("test.csv", test, fmt='%s', delimiter=',')

発生している問題・エラーメッセージ

Traceback (most recent call last):
  File "locate_probe.py", line 74, in <module>
    np.savetxt("test.csv", test, fmt='%s', delimiter=',')
  File "<__array_function__ internals>", line 6, in savetxt
  File "/opt/anaconda3/2019.10/lib/python3.7/site-packages/numpy/lib/npyio.py", line 1
    "Expected 1D or 2D array, got %dD array instead" % X.ndim)
ValueError: Expected 1D or 2D array, got 3D array instead

Answer 1

testが３次元以上の次元数を持っていると提示エラーが発生します。
savetxtは２次元以下のデータしか保存できません。CSV形式ではありませんが、代わりにnumpy.saveで保存できます。

import numpy as np

test = [[[1]]]
#np.savetxt("test.csv", test, fmt='%s', delimiter=',') # ValueError: Expected 1D or 2D array, got 3D array instead
np.save("test.npy", test) # OK