Conditionals

if runs a block only when its condition is true. elif ("else if") gives Python another condition to check when the first one fails, and else catches whatever's left over. They're checked top to bottom, and only the first matching branch runs.

Indentation is the block, in Python - there are no curly braces. Everything lined up under an if (or elif, or else) belongs to that branch; the next line back at the original indent level is outside it again.

Your task: write a function classify(n) that returns "positive" if n is greater than 0, "negative" if it's less than 0, and "zero" otherwise.

You'll practice:

Branching with if / elif / else
Returning different values from different branches

{
  "language": "python",
  "starterCode": "# Write classify(n): return \"positive\", \"negative\", or \"zero\".\ndef classify(n):\n    pass",
  "solution": "def classify(n):\n    if n > 0:\n        return \"positive\"\n    if n < 0:\n        return \"negative\"\n    return \"zero\"",
  "hints": ["Check n > 0 first and return \"positive\" if it's true.", "Check n < 0 next for \"negative\".", "Anything left over (n == 0) is \"zero\" - no condition needed for that one."],
  "tests": [
    { "name": "classify(5) is positive", "code": "assert classify(5) == \"positive\", 'classify(5) should be \"positive\"'" },
    { "name": "classify(-3) is negative", "code": "assert classify(-3) == \"negative\", 'classify(-3) should be \"negative\"'" },
    { "name": "classify(0) is zero", "code": "assert classify(0) == \"zero\", 'classify(0) should be \"zero\"'" }
  ]
}