Module 2 · Lesson 2

Boolean Indexing

Comparing an array to a value produces a boolean mask — an array of True/False the same shape as the original. Indexing with that mask returns only the elements where it’s True. To combine conditions, you must use the bitwise operators &, |, ~ (NOT Python’s and/or/not — those don’t work elementwise and will raise an error on arrays with more than one element). np.where(condition, a, b) is the vectorized if/else: pick from a where condition is True, else from b.

Two more things come up constantly once you’re working with masks. First, True/False behave as 1/0 in arithmetic, so mask.sum() counts how many elements are True — a quick way to count matches without reaching for the full aggregation toolkit (Module 4 covers .sum(), .mean(), and friends in depth). Second, indexing itself never mutates anything, but if you’re about to change values and want to leave the original untouched, call .copy() first to get an independent array — Lesson 4 covers views vs. copies in full.

arr = np.array([3, -1, 7, -5, 2, 0, 9])
mask = arr > 0
print(mask)
print(arr[mask])                      # positive values only
print(arr[(arr > 0) & (arr < 8)])     # combine with & , not "and"

clipped = np.where(arr < 0, 0, arr)   # replace negatives with 0
print(clipped)

arr[arr < 0] = 0                      # assign through a mask, in place
print(arr)

Output:

[ True False  True False  True False  True]
[3 7 2 9]
[3 7 2]
[3 0 7 0 2 0 9]
[3 0 7 0 2 0 9]

Write clip_range(arr, low, high) that returns a new array where every value below low is replaced by low, and every value above high is replaced by high (values already in range are unchanged). Do not use np.clip — build it from boolean masks and np.where (or boolean assignment on a copy) to practice the mechanics.

💡 Hint

Make a copy first (out = arr.copy()), then assign through two masks: out[out < low] = low and out[out > high] = high.

✅ Solution

def clip_range(arr, low, high):
    out = arr.copy()
    out[out < low] = low
    out[out > high] = high
    return out

Write count_in_band(arr, low, high) that returns how many elements of arr satisfy low <= x <= high, as a plain Python int.

💡 Hint

Combine two comparisons with &, then .sum() the boolean mask — True counts as 1. Wrap the result in int(...).

✅ Solution

def count_in_band(arr, low, high):
    mask = (arr >= low) & (arr <= high)
    return int(mask.sum())

Boolean Indexing

Clip to a range

Count values in a band