python simple example - ValueError: setting an array element with a sequence-CodePudding

When I execute this python code, I got an error message : ValueError: setting an array element with a sequence.

What's the problem? could you help me?

a=np.array([1,3,5,7,9], dtype=int)
c=np.array([3,4,7,8,9], dtype=int)
b=np.zeros(len(a))

for i in range(len(a)):
    b[i]= np.where(a == int(c[i]))

CodePudding user response：

The problem lies in b[i]= np.where(a == int(c[i]))this returns an array that only contains a single element in your example. However, v is also one dimensional and you have to assign scalars not arrays. If you are sure that the search only return 1 element, you can do b[i]= np.where(a == int(c[i]))[0]

CodePudding user response：

You can maybe try to append it to b instead of replacing the value?

     B.append(np.where( a == c[i]))

Also try using if conditions inside the for loop :)

CodePudding user response：

The problem lies with your array b. Since it is an array, it is not as flexible as a list. Since your np.where statement sometimes returns an empty array or even possibly an array with more scalars in it, it is a bit problematic. You will be better of to define b as a list:

a=np.array([1,3,5,7,9], dtype=int)
c=np.array([3,4,7,8,9], dtype=int)
b=[]
for i in range(len(a)):
     b.append(np.where(a==c[i])[0])

print(b)

[array([1]),
 array([], dtype=int64),
 array([3]),
 array([], dtype=int64),
 array([4])]

Only if you are entirely certain that your np.where statement will return one and only one scalar, you can use the solution of @Simon Hawe.

Do note that this returns the indexes of a where the statement is True! If you want the value itself, the code becomes

a=np.array([1,3,5,7,9], dtype=int)
c=np.array([3,4,7,8,9], dtype=int)
b=[]
for i in range(len(a)):
    b.append(a[np.where(a==c[i])])

print(b)

[array([3]),
 array([], dtype=int64),
 array([7]),
 array([], dtype=int64),
 array([9])]