for i in range(len(alist)): print(alist[i]) 1. 2. 3. 4. 5. 6. 7. enumerate()函数 enumerate - Python docs Python enumerate() 函数 | 菜鸟教程 如果你在遍历序列时,需要同时获取下标和值,可以使用enumerate()函数 for i in range(len(nums)): print(i, nums[i]) print("---") for ...
nums = [111, 222, 333, 444, 555]fornuminnums:ifnum == 333:#breakcontinueprint(num)else:print('===') 5.(1)让某段代码重复运行3次-》while循环实现如下 i =0whilei < 3:print('hello1')print('hello2')print('hello3') i+= 1 (2)让某段代码重复运行3次-》for循环实现如下 forxinra...
(nums) right = [len(nums) - 1] * len(nums) stack = [] for i, x in enumerate(nums): while len(stack) > 1 and x <= stack[-1][1]: tmp = stack.pop(-1) left[tmp[0]] = stack[-1][0] + 1 else: stack.append((i, x)) stack = [] for i in range(len(nums) - 1...
for i,x in enumerate(fruit): print(i,x) ``` **输出结果:** *0 APPLE* *1 PEAR* *2 PINEAPPLE* *3 ORANGE* *4 BANANA* Tips6、 反向,顺序遍历 将fruit列表元素从后往前依次输出,我们只需使用reversed(),如果按照a~z的顺序输出呢,pyt还是提供了一个函数:sorted(),例: ```python #将fr...
for i in nums: for e in i: print(e, end=' ') print() We have a two-dimensional list of integers. We loop over the elements with twoforloops. $ ./for_loop_nested.py 1 2 3 4 5 6 7 8 9 Python for loop with zip Thezipfunction creates an iterator from the given iterables....
for num in nums[:]: if num > 5: nums.pop(nums.index(num)) print(nums) 1. 2. 3. 4. 5. 6. 结果: [3, 3, 4, 1] while 循环语句,与if相比,若满足条件,一直运行,而不是一次。 while expression: something need to repeat 1. ...
遍历方式假设:nums=4,5,6,10,1第一种,for in的语法,这种语法很方便,但是在写Python算法里面用到的少for num in nums: print (num)第二种是下标访问,range...生成0到数组最大长度的下标数组for index in range(len(nums)): print (index,nums[index])第三种是enumerate生成索引序列序列,包含下标和元素....
astype("float32") for i, ins in enumerate(feed): nums = np.array(ins["x"]).reshape(1, 1, 13) new_data[i] = nums feed = {"x": new_data} return feed, fetch, is_batch uci_service = UciService(name="uci") uci_service.load_model_config("uci_housing_model") uci_service....
for elem in tu: print(elem) # c.请使用for、len、range输出元祖的索引 for idx in range(len(tu)): print(idx) # d.请使用enumerate输出元祖元素和序号(序号从10开始) for idx, elem in enumerate(tu, 10): print(idx, elem) tu =(
if len(x)==0: return 0 v = min(left,right) return sum([v-i for i in x]) res = 0 left = height[0] tmp = [] for i in range(1, len(height)): right = height[i] # 如果比它高,我们就计算蓄水池,然后重新链接 if right >= left: ...