[Solved] You have a numpy array z = np.array([1+1j, 0+0j, -1-1j], dtype=np.complex128).... Which code correctly does this? | NumPy

NumPy - Array Data Types

You have a numpy array z = np.array([1+1j, 0+0j, -1-1j], dtype=np.complex128). You want to create a new array that contains the magnitudes but replaces zeros with the string 'zero'. Which code correctly does this?

Anp.array([str(m) if m == 0 else m for m in np.abs(z)])

Bnp.abs(z).replace(0, 'zero')

Cnp.where(np.abs(z) == 0, 'zero', np.abs(z))

Dnp.abs(z).astype(str).replace('0.0', 'zero')

Step-by-Step Solution

Solution:

Step 1: Calculate magnitudes with np.abs
np.abs(z) returns an array of magnitudes as floats.
Step 2: Replace zeros with 'zero' using np.where
np.where(condition, value_if_true, value_if_false) replaces elements where magnitude is 0 with string 'zero', else keeps magnitude.
Final Answer:
np.where(np.abs(z) == 0, 'zero', np.abs(z)) -> Option C
Quick Check:
Use np.where for conditional replacement in arrays [OK]

Quick Trick: Use np.where(condition, 'zero', values) for replacements [OK]

Common Mistakes:

Trying to use .replace on numpy arrays (not supported)
Converting all to strings unnecessarily
Using list comprehension without numpy benefits

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Which code correctly does this?

Step 1: Calculate magnitudes with np.abs

Step 2: Replace zeros with 'zero' using np.where

Final Answer:

Quick Check:

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