在Python中连接列表列表中的元素
我有: 将其中的项目连接起来的最佳方式是什么,这样我就可以得到如下结果:在Python中连接列表列表中的元素,python,list,for-loop,Python,List,For Loop,我有: 将其中的项目连接起来的最佳方式是什么,这样我就可以得到如下结果: df1 = [['10103', 'Baldwin, C', 'SFEN'], ['10115', 'Wyatt, X', 'SFEN']] 如果我尝试像 [['10103 - Baldwin, C', 'SFEN'], ['10115 - Wyatt, X', 'SFEN']] 我得到TypeError:sequence item 0:expected str instance,list found与您的方法相同 >
df1 = [['10103', 'Baldwin, C', 'SFEN'], ['10115', 'Wyatt, X', 'SFEN']]
如果我尝试像
[['10103 - Baldwin, C', 'SFEN'], ['10115 - Wyatt, X', 'SFEN']]
我得到
TypeError:sequence item 0:expected str instance,list found
与您的方法相同
>>df1=['10103',鲍德温,C','SFEN',['10115','Wyatt,X','SFEN']]
>>>对于i,枚举中的项(df1):
... df1[i]=['-'.加入(项目[:2]),项目[2]]
...
>>>df1
['10103-Baldwin,C','SFEN',['10115-Wyatt,X','SFEN']]
还有一条路要走
>>df1=['10103',鲍德温,C','SFEN',['10115','Wyatt,X','SFEN']]
>>>df1=['-'.在df1中为item1、item2、item3连接([item1、item2]),item3]
>>>df1
['10103-Baldwin,C','SFEN',['10115-Wyatt,X','SFEN']]
join
获取字符串列表,您正在传递字符串列表
for i in df1:
df1[0:1] = [' - '.join(df1[0:1])]