将元组的部分打包成元组

I have a tuple like this

我有这样的元组

t = (1, '0076', 'AU', '9927016803A', '9927013903B', '0010', 'VO')

and I want to extract the first 6 values as tuple (in order) and the last value as a string.

我想将前6个值提取为元组（按顺序），将最后一个值提取为字符串。

The following code already works, but I am wondering if there is a "one-liner" to achieve what I'm looking for.

以下代码已经有效，但我想知道是否有一个“单行”来实现我正在寻找的东西。

# works, but it's not nice to unpack each value individually
cid,sc,ma,comp,mat,step,alt = t 
t_new = (cid,sc,ma,comp,mat,step,)
print(t_new, alt) # (1, '0076', 'AU', '9927016803A', '9927013903B', '0010') VO

This is very close to what I'm looking for, but it returns the first values as a list instead of a tuple:

这非常接近我正在寻找的东西，但它将第一个值作为列表而不是元组返回：

# works, but returns list
*t_new,alt = t 
print(t_new, alt) # [1, '0076', 'AU', '9927016803A', '9927013903B', '0010'] VO

I've already tried the following, but w/o success:

我已经尝试了以下内容，但没有成功：

tuple(*t_new),alt = t # SyntaxError
(*t_new),alt = t # still a list
(*t_new,),alt = t # ValueError

If there's no other way, I will probably go with my second attempt and cast the list to a tuple.

如果没有其他方法，我可能会进行第二次尝试并将列表转换为元组。

3 个解决方案

#1

why not just:

为什么不呢：

t = (1, '0076', 'AU', '9927016803A', '9927013903B', '0010', 'VO')

t_new, alt = t[:-1], t[-1]
print(t_new, alt)   # (1, '0076', 'AU', '9927016803A', '9927013903B', '0010') VO

#2

Either just convert it to a tuple again like you said:

要么就像你说的那样再把它转换成一个元组：

*t_new, alt = t
t_new = tuple(t_new)

Or just use slicing:

或者只使用切片：

t_new = t[:-1]  # Will be a tuple
alt = t[-1]

If you want to talk about efficiency, tuple packing / unpacking is relatively slow when compared with slicing, so the bottom one should be the fastest.

如果你想谈论效率，与切片相比，元组打包/解包相对较慢，所以底部应该是最快的。

#3

If you always want to have first 6 values in new tuple:

如果您总是希望在新元组中有前6个值：

t = (1, '0076', 'AU', '9927016803A', '9927013903B', '0010', 'VO')
newT = t[0:6]

#1