Python 从数据帧中的列表创建元组列表

Python 从数据帧中的列表创建元组列表,python,pandas,list,tuples,Python,Pandas,List,Tuples,我有这样一个数据帧: Path | Routes_tuples _______________________________________ [OAK, PHX, FLL, PBG] | 0 [OAK, SEA, FLL, PBG] | 0 [OAK, LAS, FLL, PBG] | 0 [OAK, DFW, FLL, PBG] | 0 ............. | ............... Path

我有这样一个数据帧:

Path                   |  Routes_tuples
_______________________________________

[OAK, PHX, FLL, PBG]   | 0
[OAK, SEA, FLL, PBG]   | 0
[OAK, LAS, FLL, PBG]   | 0
[OAK, DFW, FLL, PBG]   | 0
.............          | ...............
Path                   |  Routes_tuples
_______________________________________

[OAK, PHX, FLL, PBG]   | [(OAK, PHX), (PHX, FLL),(FLL, PBG)]
[OAK, SEA, FLL, PBG]   | [(OAK, SEA), (SEA, FLL),(FLL, PBG)]
[OAK, LAS, FLL, PBG]   | [(OAK, LAS), (LAS, FLL),(FLL, PBG)]
[OAK, DFW, FLL, PBG]   | [(OAK, DFW), (DFW, FLL),(FLL, PBG)]
.............          | ...............
我想要这样的东西:

Path                   |  Routes_tuples
_______________________________________

[OAK, PHX, FLL, PBG]   | 0
[OAK, SEA, FLL, PBG]   | 0
[OAK, LAS, FLL, PBG]   | 0
[OAK, DFW, FLL, PBG]   | 0
.............          | ...............
Path                   |  Routes_tuples
_______________________________________

[OAK, PHX, FLL, PBG]   | [(OAK, PHX), (PHX, FLL),(FLL, PBG)]
[OAK, SEA, FLL, PBG]   | [(OAK, SEA), (SEA, FLL),(FLL, PBG)]
[OAK, LAS, FLL, PBG]   | [(OAK, LAS), (LAS, FLL),(FLL, PBG)]
[OAK, DFW, FLL, PBG]   | [(OAK, DFW), (DFW, FLL),(FLL, PBG)]
.............          | ...............

如何使用Python实现这一点?

要获得所需内容,只需移动列表并逐步配对即可

将熊猫作为pd导入
系列=pd.系列([[OAK',PHX',FLL',PBG'],[OAK',SEA',FLL',PBG'])
couples=lambda ls:[(x,y)表示zip中的x,y(ls[:-1],ls[1:])]
系列。应用(对)
输出:

0[(橡木,PHX),(PHX,FLL),(FLL,PBG)]
1[(橡木,海),(海,FLL),(FLL,PBG)]

因此,在您的情况下,您可能只需要执行
df['Path']即可。应用(成对)

要获得您想要的,您只需移动列表并逐步配对即可

将熊猫作为pd导入
系列=pd.系列([[OAK',PHX',FLL',PBG'],[OAK',SEA',FLL',PBG'])
couples=lambda ls:[(x,y)表示zip中的x,y(ls[:-1],ls[1:])]
系列。应用(对)
输出:

0[(橡木,PHX),(PHX,FLL),(FLL,PBG)]
1[(橡木,海),(海,FLL),(FLL,PBG)]
因此,在您的情况下,您可能只需要执行
df['Path'].应用(夫妇)