How to delete an object from a numpy array without knowing the index
Question:
Is it possible to delete an object from a numpy array without knowing the index of the object but instead knowing the object itself?
I have seen that it is possible using the index of the object using the np.delete function, but I’m looking for a way to do it having the object but not its index.
Example:
[a,b,c,d,e,f]
x = e
I would like to delete x.
Answers:
You can find the index/indices of the object using np.argwhere, and then delete the object(s) using np.delete.
Example:
x = np.array([1,2,3,4,5])
index = np.argwhere(x==3)
y = np.delete(x, index)
print(x, y)
Boolean indexing or masking is a good basic way of selecting, or removing specific elements of an array
You talk about removing a specific ‘object’. Let’s take that literally and define an array of dtype object:
In [2]: x=np.array(['a','b','c','d','e'],dtype=object)
In [3]: x
Out[3]: array(['a', 'b', 'c', 'd', 'e'], dtype=object)
In [4]: x=='d' # elements that equal 'd'
Out[4]: array([False, False, False, True, False], dtype=bool)
In [5]: x!='d' # elements that don't
Out[5]: array([ True, True, True, False, True], dtype=bool)
In [6]: x[x!='d'] # select a subset
Out[6]: array(['a', 'b', 'c', 'e'], dtype=object)
Behind the scenes argwhere and delete use this. Note that argwhere uses the x==d boolean array, converting it to array indices. And constructing mask like this is one way that delete operates.
There are some important limits:
-
that equality (or not equality) test has to work for your values. It might not if the elements are floats.
-
deleting from a 1d array is easier than from a 2d (or larger) one. With 2d you have to decide whether to delete a row, a column, or an element (and in the process flattening the array).
-
deleting only one element of that matches is a bit trickier.
For some cases it might be better to .tolist() the array and use a list method.
In [32]: xl=x.tolist()
In [33]: xl.remove('d')
In [34]: np.array(xl,dtype=object)
Out[34]: array(['a', 'b', 'c', 'e'], dtype=object)
There’s no exact equivalent to list.remove for arrays.
Cast it as a numpy array, and mask it out:
x = np.array(list("abcdef"))
x = x[x!='e'] # <-- THIS IS THE METHOD
print x
# array(['a', 'b', 'c', 'd', 'f'])
Doesn’t have to be more complicated than this.
arr = np.array(['a','b','c','d','e','f'])
Then
arr = [x for x in arr if arr != 'e']
You could use np.setdiff1d(a,b), it returns all unique elements from a that are not in b.
>>> arr = np.array(['a', 'a', 'b', 'c', 'd', 'e', 'f'])
>>> to_remove = ['b', 'c']
>>> np.setdiff1d(arr, to_remove)
array(['a', 'd', 'e', 'f'], dtype='<U1')
If you have a multi-dimension np array:
arr= np.array([1,2,3,], [1,2,3,4],['a','b'])
You should do this way to remove specific values:
Create a boolean mask of values to be deleted
mask = arr == 3
Delete elements from the array using the boolean mask
new_arr = np.delete(arr, np.where(mask))
Print the new array
print(new_arr)
But in this way shape is changing. for example, if I have (12,3) ndarray it will be converted to (36,). To keep it fixed like (12,3):
Create a boolean mask of values to be deleted
mask = arr == 3
Flatten the boolean mask
flat_mask = np.ravel(mask)
Delete elements from the array using the flattened boolean mask
new_arr = np.delete(arr, np.where(flat_mask))
Reshape the new array to the original shape
new_arr = new_arr.reshape(arr.shape[0], arr.shape[1])
Print the new array
print(new_arr)
I hope it will be useful for you.
Is it possible to delete an object from a numpy array without knowing the index of the object but instead knowing the object itself?
I have seen that it is possible using the index of the object using the np.delete function, but I’m looking for a way to do it having the object but not its index.
Example:
[a,b,c,d,e,f]
x = e
I would like to delete x.
You can find the index/indices of the object using np.argwhere, and then delete the object(s) using np.delete.
Example:
x = np.array([1,2,3,4,5])
index = np.argwhere(x==3)
y = np.delete(x, index)
print(x, y)
Boolean indexing or masking is a good basic way of selecting, or removing specific elements of an array
You talk about removing a specific ‘object’. Let’s take that literally and define an array of dtype object:
In [2]: x=np.array(['a','b','c','d','e'],dtype=object)
In [3]: x
Out[3]: array(['a', 'b', 'c', 'd', 'e'], dtype=object)
In [4]: x=='d' # elements that equal 'd'
Out[4]: array([False, False, False, True, False], dtype=bool)
In [5]: x!='d' # elements that don't
Out[5]: array([ True, True, True, False, True], dtype=bool)
In [6]: x[x!='d'] # select a subset
Out[6]: array(['a', 'b', 'c', 'e'], dtype=object)
Behind the scenes argwhere and delete use this. Note that argwhere uses the x==d boolean array, converting it to array indices. And constructing mask like this is one way that delete operates.
There are some important limits:
-
that equality (or not equality) test has to work for your values. It might not if the elements are floats.
-
deleting from a 1d array is easier than from a 2d (or larger) one. With 2d you have to decide whether to delete a row, a column, or an element (and in the process flattening the array).
-
deleting only one element of that matches is a bit trickier.
For some cases it might be better to .tolist() the array and use a list method.
In [32]: xl=x.tolist()
In [33]: xl.remove('d')
In [34]: np.array(xl,dtype=object)
Out[34]: array(['a', 'b', 'c', 'e'], dtype=object)
There’s no exact equivalent to list.remove for arrays.
Cast it as a numpy array, and mask it out:
x = np.array(list("abcdef"))
x = x[x!='e'] # <-- THIS IS THE METHOD
print x
# array(['a', 'b', 'c', 'd', 'f'])
Doesn’t have to be more complicated than this.
arr = np.array(['a','b','c','d','e','f'])
Then
arr = [x for x in arr if arr != 'e']
You could use np.setdiff1d(a,b), it returns all unique elements from a that are not in b.
>>> arr = np.array(['a', 'a', 'b', 'c', 'd', 'e', 'f'])
>>> to_remove = ['b', 'c']
>>> np.setdiff1d(arr, to_remove)
array(['a', 'd', 'e', 'f'], dtype='<U1')
If you have a multi-dimension np array:
arr= np.array([1,2,3,], [1,2,3,4],['a','b'])
You should do this way to remove specific values:
Create a boolean mask of values to be deleted
mask = arr == 3
Delete elements from the array using the boolean mask
new_arr = np.delete(arr, np.where(mask))
Print the new array
print(new_arr)
But in this way shape is changing. for example, if I have (12,3) ndarray it will be converted to (36,). To keep it fixed like (12,3):
Create a boolean mask of values to be deleted
mask = arr == 3
Flatten the boolean mask
flat_mask = np.ravel(mask)
Delete elements from the array using the flattened boolean mask
new_arr = np.delete(arr, np.where(flat_mask))
Reshape the new array to the original shape
new_arr = new_arr.reshape(arr.shape[0], arr.shape[1])
Print the new array
print(new_arr)
I hope it will be useful for you.