将两个元素结合在一起的列表理解

Question

如何通过将元素1连接到元素2等将lst1转换为lst2？

lst1=[' ff 55 00 90 00 92 00 ad 00 c6 00 b7 00 8d 00 98 00 87 00 8a 00 98 00 8f 00 ca 01 78 03 54 05 bf']

到

lst2=[ff55, 0090, 0092, 00ad, 00c6, 00b7, 008d, 0098, 0087, 008a, 0098, 008f, 00ca, 0178, 0354,05bf]

我试过了，但并不像预期的那样：

   for i in lst:
        lstNew = []
        tempList =  i.split()
        lenList = len(tempList)
        #print tempList
        index = 0
        while (index < lenList):
            print tempList[index] + tempList[index+1]
            index = index + 2

Answer 1

这样可以吗：

>>> lst = ['ff', '55', '00', '90', '00', '92', '00', 'ad', 
           '00', 'c6', '00', 'b7', '00', '8d', '00', '98', 
           '00', '87', '00', '8a', '00', '98', '00', '8f', 
           '00', 'ca', '01', '78', '03', '54', '05', 'bf']

>>> [ ''.join(x) for x in zip(lst[0::2], lst[1::2]) ]
    ['ff55', '0090', '0092', '00ad', '00c6', '00b7', '008d', 
     '0098', '0087', '008a', '0098', '008f', '00ca', '0178', 
     '0354', '05bf']
>>>

或者

>>> [ x+y for x,y in zip(lst[0::2], lst[1::2]) ]
['ff55', '0090', '0092', '00ad', '00c6', '00b7', 
 '008d', '0098', '0087', '008a', '0098', '008f', 
 '00ca', '0178', '0354', '05bf']  
>>>

Answer 2

假设您有一个由空格分隔的两个字母组成的字符串，则可以执行以下操作：

>>> from funcy import ichunks
>>> s = "ff 55 00 90 00 92 00 ad 00 c6 00 b7 00 8d 00 98 00 87 00 8a 00 98 00 8f 00 ca 01 78 03 54 05 bf"
>>> ["".join(chunk) for chunk in ichunks(2, s.split())]
['ff55', '0090', '0092', '00ad', '00c6', '00b7', '008d', '0098', '0087', '008a', '0098', '008f', '00ca', '0178', '0354', '05bf']
>>>

这利用了我最喜欢的库funcy

更新：

正如评论中所建议的；这里有一个稍微好一点的版本：

>>> from funcy import ichunks
>>> s = "ff 55 00 90 00 92 00 ad 00 c6 00 b7 00 8d 00 98 00 87 00 8a 00 98 00 8f 00 ca 01 78 03 54 05 bf"
>>> [c[:2] + c[3:5] for c in ichunks(6, s)]
['ff55', '0090', '0092', '00ad', '00c6', '00b7', '008d', '0098', '0087', '008a', '0098', '008f', '00ca', '0178', '0354', '05bf']
>>>

Answer 3

你的清单格式是

lst1=[' ff 55 00 90 00 92 00 ad 00 c6 00 b7 00 8d 00 98 00 87 00 8a 00 98 00 8f 00 ca 01 78 03 54 05 bf']

让我们替换所有空格并将其转换为字符串

list1=''.join([i.replace(" ","") for i in lst1])

现在我们可以每4个字符递增一次得到结果

list1= [list1[i:i+4]for i in range(0,len(list1),4)]
print list

#output=['ff55', '0090', '0092', '00ad', '00c6', '00b7', '008d', '0098', '0087', '008a', '0098', '008f', '00ca', '0178', '0354', '05bf']