我有2个数组，其中一个是Large_arr，另一个是sub_arr（是它的一个子集）。我要映射数组的值，并且数组的匹配值将返回大型数组的索引。我有我的代码，但抛出错误
Traceback (most recent call last):
File "C:/Users/hp/AppData/Local/Programs/Python/Python36/stack.py", line 36, in <module>
if(sub_arr[i]==Large_arr[j]):
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
我的代码是
import numpy as np
Large_arr=np.array([[0.12046862, 0.64457892],
[0.47522597, 0.12350968],
[0.1649585 , 0.50135502],
[0.20104755 ,0.15218623],
[0.03772419 ,0.42482297],
[0.51633778 ,0.61209002],
[0.24848648 ,0.99651906],
[0.47374345, 0.09990318],
[0.58390815 ,0.19781604],
[0.9613725 ,0.45975827],
[0.99008266 ,0.13487207],
[0.14410988 ,0.36196475],
[0.81349573 ,0.55896232],
[0.72841399 ,0.02263056],
[0.8692731 ,0.9754183 ],
[0.87142787 ,0.66163271],
[0.24342035 ,0.95821073],
[0.94218857 ,0.7220602 ],
[0.66716105 ,0.96875209]])
sub_arr=np.array([[0.12046862, 0.64457892],
[0.51633778 ,0.61209002],
[0.99008266 ,0.13487207],
[0.72841399 ,0.02263056],
[0.24342035 ,0.95821073],
[0.47374345, 0.09990318],
[0.9613725 ,0.45975827]])
s=[]
for i in range(0,len(Large_arr)):
for j in range(0,len(sub_arr)):
if(sub_arr[i]==Large_arr[j]):
s.append(j)
print("Value of s is\n",s)
else:
print("Value is none\n")
可以简化此方法吗？所以我的输出很有价值
S=[0,5,10,13,7,10,1] (example, index of large array where sub_array value stored)
答案 0 :(得分：0)
首先，在循环中交换i和j索引。确保将它们用于索引到相应的数组。
第二，您正在执行的代码将子数组按元素进行比较，这将返回一个布尔数组，这是一个数组，而不是布尔本身，因此必须添加.all（）以将其简化为单个布尔值。如果所有元素都为true，则.all（）将为true，否则为false。
第三，您应该可以减少它。我在下面包含了代码，以显示所有这三点。
此外，您的换行符没有执行应有的操作。请确保在发布代码之前先对其进行测试，并提供详细信息，例如所使用的python版本。
import numpy as np
Large_arr=np.array([[0.12046862, 0.64457892],
[0.47522597, 0.12350968],
[0.1649585 , 0.50135502],
[0.20104755 ,0.15218623],
[0.03772419 ,0.42482297],
[0.51633778 ,0.61209002],
[0.24848648 ,0.99651906],
[0.47374345, 0.09990318],
[0.58390815 ,0.19781604],
[0.9613725 ,0.45975827],
[0.99008266 ,0.13487207],
[0.14410988 ,0.36196475],
[0.81349573 ,0.55896232],
[0.72841399 ,0.02263056],
[0.8692731 ,0.9754183 ],
[0.87142787 ,0.66163271],
[0.24342035 ,0.95821073],
[0.94218857 ,0.7220602 ],
[0.66716105 ,0.96875209]])
sub_arr=np.array([[0.12046862, 0.64457892],
[0.51633778 ,0.61209002],
[0.99008266 ,0.13487207],
[0.72841399 ,0.02263056],
[0.24342035 ,0.95821073],
[0.47374345, 0.09990318],
[0.9613725 ,0.45975827]])
s=[]
for i in range(0,len(Large_arr)):
for j in range(0,len(sub_arr)):
if((sub_arr[j]==Large_arr[i]).all()):
s.append(i)
print("Value of s is\n",s)
else:
#print("Value is none\n")
pass
opt = np.isin(Large_arr,sub_arr)
print
print "Bool array"
print opt
print
print "indexes"
print np.where(opt[:,0])[0]
答案 1 :(得分：0)
您可以尝试以下方法：
result_indices = np.argwhere(np.in1d(Large_arr, sub_arr))
那应该首先应用一个布尔掩码，然后为您提供匹配元素的索引。
如果这最终不是您想要的，请纠正我。
答案 2 :(得分：0)
我认为问题可能出在您的
if(sub_arr[i]==Large_arr[j]):
实际上，您似乎有两个二维数组，因此不能仅指定一个。