在这里,我们将使用外部 for 循环迭代列表。 外层for循环每次迭代,内层for循环执行5次,打印当前名称5次。 names = ['Kelly', 'Jessa', 'Emma'] # outer loop for name in names: # inner while loop count = 0 while count < 5: print(name, end=' ') # print(name) # increment counter count = ...
if self.counter >= 10: raise StopAsyncIteration # increment the counter self.counter += 1 # simulate work await asyncio.sleep(1) # return the counter value return self.counter # main coroutine async def main(): # loop over async iterator with async for loop async for item in AsyncIterat...
// Counter-controlled repetition with the for statement #include <iostream> using std::cout; using std::endl; int main() { // for statement header includes initialization // loop-continuation condition and increment for( int counter = 1; counter <= 10; counter++ ) cout << counter << "...
def counter(): count = 0 def increment(): nonlocal count count += 1 return count return increment counter_func = counter() print(counter_func()) # 输出: 1 print(counter_func()) # 输出: 22.1.2 递归与匿名函数(lambda表达式) 递归是函数直接或间接地调用自身的过程,常用于处理分治问题。例如,...
list using for loop for day in range(len(weekdays)):print(weekdays[day])while循环 # Initialize counter counter = 1 # Inerate the loop 5 times while counter < 6:#print the counter value print("The current counter value:%d" % counter)# Increment the counter counter = counter + 1 ...
//Counter-controlled repetition with the for statement#include<iostream>usingstd::cout;usingstd::endl;intmain() {//for statement header includes initialization//loop-continuation condition and incrementfor(intcounter =1; counter <=10; counter++) ...
In each iteration of outer for loop, the inner for loop execute five times to print the current name five times names = ['Kelly','Jessa','Emma']# outer loopfornameinnames:# inner while loopcount =0whilecount <5: print(name, end=' ')# increment countercount = count +1print() ...
current counter value: %d" % counter) # Increment the counter counter = counter + 15...
如果你想重复x次,你可以使用一个循环for _ in range(x),它对_本身没有任何作用。 将任务解读为 使用for循环将start中的count值与increment相加 这意味着count不是范围的上限——你就是这么做的——而是要加起来的值的数量。因此,对于count次,将start值增加increment,并取值之和。 def sum_count(start, increme...
counter=1# Iterate the loop5timeswhilecounter<6:# Print the counter valueprint("The current counter value: %d"%counter)# Increment the counter counter=counter+1 「5、import导入其他脚本的功能」 有时需要使用另一个 python 文件中的脚本,这其实很简单,就像使用 import 关键字导入任何模块一样。