Python新手求助：详解加密题解法中exec语句的s[x::{0}]与range({0})

Hey there! Let's break down this code step by step—it's a clever (though unnecessarily compact) solution for the Encryption problem. Let's unpack everything, starting with the exec statement and the parts you're curious about.

First, What's the Big Picture? #

The Encryption problem asks you to rearrange a string into a grid (with rows and columns calculated from the string's length) and then read the string column-by-column, separating each column's result with a space. For example, if the input is "helloworld", the grid would be 3 rows × 4 columns, and the encrypted output is "hol ewd lo lr".

Let's Demystify the exec Statement #

First, exec() is a Python built-in that runs a string as if it were actual Python code. The string here uses .format() to plug in a value: ceil(sqrt(len(s))). Let's call this value cols (it's the number of columns we need for the encryption grid).

After replacing {0} with cols, the code inside exec becomes:

print(' '.join(map(lambda x: s[x::cols], range(cols))))

That's way easier to read! The exec here is just a fancy (and confusing) way to write this line directly—there's no practical reason to use exec here; it's just code golf.

What Does range({0}) Mean? #

When {0} is replaced with cols, range(cols) generates a sequence of integers from 0 to cols-1. For example, if cols is 4, this gives [0, 1, 2, 3]. Each number in this sequence represents the starting index of a column in our encryption grid. We loop through each of these to build our encrypted string.

What About s[x::{0}]? #

This is Python's slice syntax, which follows the pattern s[start::step]. Here:

• x is the starting index (pulled from our range(cols) sequence)
• {0} is replaced with cols, which acts as the step value

So s[x::cols] takes every cols-th character starting at index x. Let's use an example to make this concrete:

• Suppose s = "helloworld" (length 10), so cols = ceil(sqrt(10)) = 4
• For x=0: s[0::4] takes indices 0, 4, 8 → characters h, o, l → "hol"
• For x=1: s[1::4] takes indices 1, 5, 9 → e, w, d → "ewd"
• For x=2: s[2::4] takes indices 2, 6 → l, o → "lo"
• For x=3: s[3::4] takes indices 3, 7 → l, r → "lr"

Putting It All Together #

• map(lambda x: s[x::cols], range(cols)) applies our slice operation to every starting index in range(cols), generating each column's string.
• ' '.join(...) combines these column strings into one, separated by spaces.
• print() outputs the final encrypted string.

A Simpler Alternative (No exec Needed!) #

This code uses exec for no real reason—you can write the same logic much more clearly without it:

#!/bin/python3
import sys
from math import ceil, sqrt

def encryption(s):
    cols = ceil(sqrt(len(s)))
    print(' '.join(s[x::cols] for x in range(cols)))

内容的提问来源于stack exchange，提问作者pythonaddict