合并两个Python元组列表

Question

I have two lists of tuples. Each tuple contains a datetime object and a float object ie:

l1 = [(dt1), 1.0), (dt2, 2.0), (dt3, 3.0)]
l2 = [(dt1), 1.0), (dt3, 3.0), (dt5, 6.0)]

The two lists are already sorted by the datetime of each tuple.

What would be a fast way to merge the two lists ?

The two lists may not contain the same datetimes. If one dt exists in one list but not the other the missing value can be ''

So for example using the two lists above I would like to produce

l = [(dt1), 1.0, 1.0), (dt2, 2.0, ''), (dt3, 3.0, 3.0), (dt5, '', 6.0)]

I'm guessing maybe I should use a dictionary with the dt as key and then resort but that seems wasteful.

Any other ideas ?

Thanks

Answer 1

l1 = [("a", 1.0), ("b", 2.0), ("c", 3.0)]
l2 = [("a", 1.0), ("c", 3.0), ("d", 6.0)]


i1 = 0
i2= 0
l = []
while i1 != len(l1) or i2!=len(l2):
    print i1,i2
    if ((i1<len(l1))and(i2<len(l2))) and l1[i1] ==l2[i2]:
        l.append((l1[i1][0],l1[i1][1],l2[i2][1]))
        i1 +=1
        i2+=1
    elif ((i1<len(l1)and i2<len(l2)) and l1[i1] <l2[i2]) or (i1<len(l1)and i2>=len(l2)):
        l.append((l1[i1][0],l1[i1][1],""))
        i1 +=1

    elif ((i2<len(l2)and i1<len(l1)) and l1[i1] >l2[i2])or (i1>=len(l1)and i2<len(l2)):
        l.append((l2[i2][0],l2[i2][1],""))
        i2 +=1    

I think I fixed the edge cases now too.

Answer 2

In case it's useful, I used a different approach:

result = []
d2 = dict(l2)
for i in l1:
    j = ''
    if i[0] in d2:
        j = d2[i[0]]
        d2.pop(i[0], None)
    result.append((i[0], i[1], j))

for key in d2:
    result.append((key, '', d2[key]))

Answer 3

Mine is about the same efficiency as Moe's answer. Notice the second value in the tuple in the second list is always placed as the third element in the tuple in the merged list

Result is Merge_List(l1,l2) , where Merge_List is:

def Merge_List(Left,Right,Merged=[]):
    if len(Left)==0 and len(Right)==1:
        return Merged+[(Right[0][0],"",Right[0][1]),]
    elif len(Right)==0 and len(Left)==1:
        return Merged+[Left[0]+("",),]
    else:
        if Left[0][0] < Right[0][0]:
            return Merge_List(Left[1:],Right,
                           Merged=Merged+[Left[0]+("",),])
        elif Right[0][0] < Left[0][0]:
            return Merge_list(Left,Right[1:],
                           Merged=Merged+[(Right[0][0],"",Right[0][1]),])
        else:
            return Merge_List(Left[1:],Right[1:],
                           Merged=Merged+[(Left[0][0],Left[0][1],Right[0][1]),])