The problem of "fruits" also can be solved in the way below:d = {} for item in fruits: d[item] = d.___________(item, 0) + 1

程序运行结果为（______ ）fruit = ['apple', 'banana', 'mango', 0]for item in fruit: if item == 0: fruit[item] = 'watermelon' elif item == 'mango': fruit[2] = 'pear'print(fruit)

有列表定义如下：[br][/br]group1 = [ ["萧峰", 98], ["杨过", 96] ] 要提取两位大侠的名字"萧峰"和"杨过"，应该采用下面哪种方法 A: n = [ item[0] for item in group1 ] B: n =[ item[1] for item in group1 ] C: n = group1[0],group1[1] D: n = [ item for item in group1 ]

已知 x = [[1,3,3],[2,3,1]]，那么表达式 sorted(x, key=lambda item:item[2]) 的值为____________________________________。

以下程序执行结果是________。 string = ["PHP", "Python", "C", "C++", "Java"] for item in string: item_temp = item.lower() for i in range(len(item)): if item_temp[i] in 'aeiou': string.remove(item) break print(string) s = sorted(string, reverse = True) print(s)[/i]

开启第二条生产线的标准（） A: 早餐ITEM数≥280/午餐ITEM≥241 B: 早餐ITEM数≥281/午餐ITEM≥240 C: 早餐ITEM数≥241/午餐ITEM≥280 D: 早餐ITEM数≥200/午餐ITEM≥180

x = ['aaaa', 'bc', 'd', 'b', 'ba'] sorted(x, key=lambda item: (len(item), item))运行结果正确的是( )。 A: ['b', 'd', 'ba', 'bc', 'aaaa'] B: ['aaaa', 'ba', 'bc', 'b', 'd'] C: ['aaaa', 'b', 'd', 'ba', 'bc'] D: ['ba', 'bc', 'aaaa', 'b', 'd']

有如下代码:<;view wx:for='{{array}}'>; {{index}}:{{item}}<;/view>; Page({ data:{ array:&#91; 1,2,3,4&#93; }}) 显示结果为:____

The opportunity cost of an item is ( ). A: the price of that item B: what you give up to get that item C: the time you have to spend on earning enough money to buy that item D: the cost of producing that item

A credit item is an item for which a country must pay. ( )