11.3 可变嵌套循环
现在来尝试一个可变嵌套循环。这就是一个嵌套循环,只不过其中一个或多个循环在 range() 函数中使用了变量。代码清单 11-2 给出了一个例子。
代码清单 11-2 一个可变嵌套循环
numLines = int(raw_input ('How many lines of stars do you want? '))
numStars = int(raw_input ('How many stars per line? '))
for line in range(0, numLines):
for star in range(0, numStars):
print '*',
运行这个程序来看它的作用,你会看到类似这样的结果:
>>> ============================ RESTART ============================
>>>
How many lines of stars do you want? 3
How many stars per line? 5
*
*
*
前两行询问用户想要多少行,以及每行希望有多少个星号。程序使用变量 numLines 和 numStars 记住这些答案。接下来有两个循环:
内循环(for star in range (0, numStars):)打印每个星号,对每一行上的每个星号分别运行一次;
外循环(for line in range (0, numLines):)对每行星号分别运行一次。
需要用第二个 print 命令开始新的一行星号。如果没有这个命令,由于第一个 print 语句中有逗号,所有星号都会打印到同一行上。
甚至可以有“嵌套嵌套循环”(或双重嵌套循环),就像代码清单 11-3 这样。
代码清单 11-3 利用双重嵌套循环生成星号块
numBlocks = int(raw_input ('How many blocks of stars do you want? '))
numLines = int(raw_input ('How many lines in each block? '))
numStars = int(raw_input ('How many stars per line? '))
for block in range(0, numBlocks):
for line in range(0, numLines):
for star in range(0, numStars):
print '*',
会得到下面的输出:
>>> ======================= RESTART =======================
>>>
How many blocks of stars do you want? 3
How many lines of stars in each block? 4
How many stars per line? 8
我们称这个循环嵌套“深度为 3”。
