2017-07-07 52 views
0

鉴于此输入:的Python - 拆分成列表列出了许多在for循环中

['1636 19 02 2F', '1637 19 02 2F', '1631 19 02 2F'] 

我想产生这样的输出:

['1636'] 
['19'] 
['02'] 
['2F'] 
['1637'] 
['19'] 
['02'] 
['2F'] 

我想:

byte = ['1636 19 02 2F', '1637 19 02 2F', '1631 19 02 2F'] 

for i in byte: 
    d = i.split(' ') 
    list_byte = d # ('\n'.join(d)) 
    print(list(list_byte,)) 

但是它的输出是:

['1636', '19', '02', '2F'] 
['1637', '19', '02', '2F'] 

我也尝试过nested loop,但它分裂成每个字符和join,但只是写出了错误的类型。

如何在单独的列表中输出这样的结果?

['1636'] 
['19'] 
['02'] 
['2F'] 
['1637'] 
['19'] 
['02'] 
['2F'] 
+0

你为什么要打印的为*单*名单? –

+1

尝试'[[我为我在j.split('')为j在字节]' –

+0

我写他们到一个Access 2016与此cursor.executemany(“”“INSERT INTO tblMsgBytes(ByteValue)VALUES(? )“”“,list_byte)然后我必须以这种格式,有一个更好的方法...但这是我可以:) –

回答

0

可能会有所帮助。

byte = ['1636 19 02 2F', '1637 19 02 2F', '1631 19 02 2F'] 

for temp in byte : 
    for list in temp.split(' '): 
     list = [list] 

输出:

['1636'] 
['19'] 
['02'] 
['2F'] 
['1637'] 
['19'] 
['02'] 
['2F'] 
['1631'] 
['19'] 
['02'] 
['2F'] 
+0

谢谢,完全按照我想要的那样工作! –

0
byte = ['1636 19 02 2F', '1637 19 02 2F', '1631 19 02 2F'] 

for a in byte: 
    a_list = a.split(' ') 
#  print(a_list) 
    for b in a_list: 
     print([b]) 

给你。

['1636'] 
['19'] 
['02'] 
['2F'] 
['1637'] 
['19'] 
['02'] 
['2F'] 
['1631'] 
['19'] 
['02'] 
['2F'] 
+0

谢谢你,完美! –

+0

很伤心QQ 首先回答但未被接受。 –

0
for a_string in byte: 
    for split_result in a_string.split(' '): 
     print [split_result] 

这给你的格式打印...

['1636'] 
['19'] 
['02'] 
['2F'] 
['1637'] 
['19'] 
['02'] 
['2F'] 
['1631'] 
['19'] 
['02'] 
['2F'] 
+0

感谢您的回答! –