Universal functions - use Technical Quotient

Ufunc methods

1. Remember

Which method on a scalar ufunc performs the operation across one or more axes of an array?

reduceat

Incorrect; reduceat performs a local reduce with specified slices.

reduce

Correct; reduce applies a ufunc operation across axes of an array to reduce its dimension.

outer

Incorrect; outer applies the ufunc on all pairs of two arrays' elements, not reducing dimensions.

accumulate

Incorrect; accumulate applies a ufunc but maintains the array's dimension.

2. Understand

What is the result of using the out parameter in a ufunc reduction operation?

The result is stored in the provided output array, potentially altering its values.

Correct; the out parameter allows the operation's result to be stored directly in a provided array, overwriting its values.

The operation is applied element-wise instead of as a reduction.

Incorrect; the operation's nature (reduction) remains unchanged by using out.

The out parameter specifies the axis to reduce over.

Incorrect; the axis parameter specifies the reduction axis, not out.

The out parameter increases the computation speed of the operation.

Incorrect; out does not affect the speed but specifies where to store the result.

3. Apply

You have an array np.array([1, 2, 3, 4]) and want to use np.add.reduce() with an output array of np.zeros(1, dtype=int). How would you write this code?\n

x = np.array([1, 2, 3, 4])\nresult = np.zeros(1, dtype=int)\nnp.add.reduce(x, axis=0)

x = np.array([1, 2, 3, 4])\nresult = np.add.accumulate(x)\nnp.add.reduce(result)

Incorrect; this code uses accumulate and then attempts to reduce the accumulated result, which is not the described operation.

x = np.array([1, 2, 3, 4])\nresult = np.add.reduce(x, axis=0)\nprint(result)

Incorrect; while this code does a reduction, it doesn't use an out parameter to store the result.

x = np.array([1, 2, 3, 4])\nresult = np.zeros(1, dtype=int)\nnp.add.reduce(x, out=result)

4. Apply

How to perform a reduction on np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) using np.multiply.reduce with dtype=float?

x = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])\nresult = np.multiply.reduce(x, dtype=float)\nprint(result)

x = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])\nresult = np.multiply.accumulate(x, dtype=float)\nprint(result)

Incorrect; accumulate is used instead of reduce, which maintains the original array's shape.

x = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])\nresult = np.add.reduce(x, axis=0)\nprint(result)

Incorrect; uses add.reduce instead of multiply.reduce, altering the operation type.

x = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])\nresult = np.zeros(3, dtype=float)\nnp.multiply.reduce(x, out=result)

5. Analyze

While attempting to reduce a large integer array using np.add.reduce without specifying dtype, you notice the output seems incorrect due to overflow. How should you modify your code to ensure accurate results?

Specify dtype=np.int64 or dtype=np.float64 to handle larger sums.
Use np.add.accumulate instead of np.add.reduce to prevent overflow.

Incorrect; accumulate maintains the array's shape and does not prevent overflow in the sum.

Split the array into smaller chunks and use np.add.reduce separately on each.
Increase the size of the input array to accommodate larger numbers.

1. Remember

2. Understand

3. Apply

4. Apply

5. Analyze

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