Python新手,这里是Python迭代
Python新手-你能解释一下吗Python新手,这里是Python迭代,python,string,iteration,Python,String,Iteration,Python新手-你能解释一下吗 greeting = 'Hello!' count = 0 for letter in greeting: count += 1 if count % 2 == 0: print(letter) print(letter) print('done') 为了避免缩进错误,我将解释代码的作用 在您的程序中,您已经初始化了一个名为greeting的变量,该变量的值为“Hello!”,并且还初始化了一个值为0的
greeting = 'Hello!'
count = 0
for letter in greeting:
count += 1
if count % 2 == 0:
print(letter)
print(letter)
print('done')
为了避免缩进错误,我将解释代码的作用 在您的程序中,您已经初始化了一个名为greeting的变量,该变量的值为“Hello!”,并且还初始化了一个值为0的计数 问候语=‘你好!’ 计数=0 此后,for-us使用了一个循环,该循环贯穿问候语,即直到每个单词Hello。但是,如果你想自己检查,你可以打印这封信 问候信: 印刷品(信件) 现在来看看您的问题,您还将count的值增加了1,其中每次循环执行时该值增加1 然后,您有一个条件来检查数字是否为偶数计数%2==0,然后是条件成功时执行的print语句。这意味着处于偶数位置的字母只能打印出来
这就是你的程序所做的。
greeting='Hello!'<代码>问候语=‘你好!’有相关的教程。该代码将不会运行,因为它充满了缩进错误。正确的缩进在Python中非常重要。请先尝试更正缩进。将程序保存到hello.py中,然后运行python hello.py检查结果。在这之后,您可以使用您的代码来理解它的行为。
greeting = 'Hello!' <-- set greeting variable.
count = 0 <-- set count variable.
for letter in greeting: <-- this loop will loop six times because greeting contains of six character.
count += 1 <-- each times of loop will increase value of count by one.
if count % 2 == 0: <-- this line will print a index of character that % 2 = 0 (eg. 2%2 = 0, 4%2 = 0, ...)
print(letter)
print(letter) <-- this line will print any index of character of greeting. (ps. this line will error because indentation errors.)
print('done') <-- print 'done'.
greeting = 'Hello!' <-- set greeting variable.
count = 0 <-- set count variable.
for letter in greeting: <-- this loop will loop six times because greeting contains of six character.
count += 1 <-- each times of loop will increase value of count by one.
if count % 2 == 0: <-- this line will print a index of character that % 2 = 0 (eg. 2%2 = 0, 4%2 = 0, ...)
print(letter)
print(letter) <-- this line will print any index of character of greeting. (ps. this line will error because indentation errors.)
print('done') <-- print 'done'.