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Predict the output: negative indexes and chained slices

Q: Why does the stop index get excluded?

So that s[:i] + s[i:] always rebuilds the original and len(s[a:b]) is b - a. The empty slice s[3:3] is the same consistency rule.

Q: Do lists behave differently from strings here?

No — indexing and slicing are identical for str, list and tuple. One difference: strings are immutable, so s[0] = "P" raises TypeError while a list allows it.

🔮 Predict the outputTopic: stringsDifficulty: easyNo login needed

The snippet

Every answer here follows from two rules — indexes count from -len, and a slice excludes its stop. The hard part is that [::-1] flips the meaning of the other two numbers, and here it is chained with a second slice.

s = "python"
print(s[1:-1], s[::-1][:2], s[-4:-2])

Your prediction

Type exactly what this program prints — one line per print() call:

The output

ytho no th

Verified against CPython 3.11 in this repo's build

Line-by-line explanation

s = "python"
Six characters: indexes 0–5, or -6 to -1.
s[1:-1]
Start at 1 (y), stop before -1 (the n) → ytho. A negative stop counts back from the end; it does not mean "go backwards".
s[::-1][:2]
[::-1] is the whole string reversed → "nohtyp"; taking [:2] of *that result* gives no. Slices produce new strings, so chaining works left to right.
s[-4:-2]
-4 is t, -2 is o; the stop is excluded → th.
printed together
print separates arguments with a space → ytho no th.

The rule you can reuse

s[a:b] includes a, excludes b; a negative index means len + index. Slicing out-of-range never raises — s[:100] just stops at the end, while s[100] raises IndexError.

Why most predictions break here

Reading s[1:-1] backwards, or expecting s[::-1][:2] to give yt. And s[2:2] is the empty string, not an error.

Quick fire

Same rule, two more shapes. Predict first, then open.

Predict this too

print("debug"[2], "debug"[:1], "debug"[-3:])

Output
b d bug

One character, a one-character prefix, and the last three → b d bug.

Predict this too

print(list(range(10))[::3])
print(list(range(10))[::-1][:2])

Output
[0, 3, 6, 9] [9, 8]

Every third item from 0 → [0, 3, 6, 9]; reversed then two → [9, 8].

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Questions this page answers

Why does the stop index get excluded?

So that s[:i] + s[i:] always rebuilds the original and len(s[a:b]) is b - a. The empty slice s[3:3] is the same consistency rule.

Do lists behave differently from strings here?

No — indexing and slicing are identical for str, list and tuple. One difference: strings are immutable, so s[0] = "P" raises TypeError while a list allows it.