相关疑难解决方法(0)

我正在创建一个python程序.

我有一个清单:

[3, 28, 25, 126, 25, 127]

如何将其转换为元组列表,以便列表变为:

[(3,28),(25,126),(25,127)]

它加入了2个元素并构成了一个元组.

我有一个列表,我想将此列表转换为地图

mylist = ["a",1,"b",2,"c",3]

mylist相当于

mylist = [Key,Value,Key,Value,Key,Value]

所以输入:

mylist = ["a",1,"b",2,"c",3]

输出:

mymap = {"a":1,"b":2,"c":3}

PS:我已经编写了以下函数来完成同样的工作,但我想使用python的迭代器工具:

def fun():
    mylist = ["a",1,"b",2,"c",3]
    mymap={}
    count = 0
    for value in mylist:
        if not count%2:
            mymap[value] = mylist[count+1]
        count = count+1
    return mymap        

这是我的清单

myList = ['Bob', '5-10', 170, 'Tom', '5-5', 145, 'Bill', '6-5', '215']

我想写成这样的字典.

{'Bob': ['5-10', 170], 'Bill': ['6-5', '215'], 'Tom': ['5-5', 145]}

我想出了这个,但它非常难看,不会扩大规模.

def MakeDict():
    d = {}
    for x, i  in zip(myList, range(len(myList))):
       if i in (0, 3, 6):
           name = x
       elif i in (1, 4, 7):
           hieght = x
       elif i in (2, 5, 8):
           wieght = x
          d[name] = [hieght, wieght]
return d

我能做什么？

您好我是python和regex的新手.我有一个大的CSV文件,其中%age composition包含类似的字段,其中包含以下值:

'34% passed 23% failed 46% deferred'

如何拆分此字符串以便获取字典对象:

{'passed': 34, 'failed': 23, 'deferred': 46} 每排？

我试过这个:

for line in csv_lines:
    for match in re.findall('[\d\s%%]*\s', line)

但这只取%年龄值

例如,我有这样的简单列表:

l = [1,2,3,4,5,6,7,8,9]

我想以3个元素为一组迭代它:

1,2,3
4,5,6
7,8,9

最简单的方法是什么？