根据每个子列表中的第三项删除列表中的重复项

Question

I have a list of lists that looks like:

c = [['470', '4189.0', 'asdfgw', 'fds'],
     ['470', '4189.0', 'qwer', 'fds'],
     ['470', '4189.0', 'qwer', 'dsfs fdv'] 
      ...]

c has about 30,000 interior lists. What I'd like to do is eliminate duplicates based on the 4th item on each interior list. So the list of lists above would look like:

c = [['470', '4189.0', 'asdfgw', 'fds'],['470', '4189.0', 'qwer', 'dsfs fdv'] ...]

Here is what I have so far:

d = [] #list that will contain condensed c
d.append(c[0]) #append first element, so I can compare lists
for bact in c: #c is my list of lists with 30,000 interior list
    for items in d:
        if bact[3] != items[3]:
            d.append(bact)  

I think this should work, but it just runs and runs. I let it run for 30 minutes, then killed it. I don't think the program should take so long, so I'm guessing there is something wrong with my logic.

I have a feeling that creating a whole new list of lists is pretty stupid. Any help would be much appreciated, and please feel free to nitpick as I am learning. Also please correct my vocabulary if it is incorrect.

Answer 1

I'd do it like this:

seen = set()
cond = [x for x in c if x[3] not in seen and not seen.add(x[3])]

Explanation:

seen is a set which keeps track of already encountered fourth elements of each sublist. cond is the condensed list. In case x[3] (where x is a sublist in c ) is not in seen , x will be added to cond and x[3] will be added to seen .

seen.add(x[3]) will return None , so not seen.add(x[3]) will always be True , but that part will only be evaluated if x[3] not in seen is True since Python uses short circuit evaluation. If the second condition gets evaluated, it will always return True and have the side effect of adding x[3] to seen . Here's another example of what's happening ( print returns None and has the "side-effect" of printing something):

>>> False and not print('hi')
False
>>> True and not print('hi')
hi
True

Answer 2

You have a significant logic flaw in your current code:

for items in d:
    if bact[3] != items[3]:
        d.append(bact)  

this adds bact to d once for every item in d that doesn't match . For a minimal fix, you need to switch to:

for items in d:
    if bact[3] == items[3]:
        break
else:
    d.append(bact)  

to add bact once if all items in d don't match. I suspect this will mean your code runs in more sensible time.

---

On top of that, one obvious performance improvement (speed boost, albeit at the cost of memory usage) would be to keep a set of fourth elements you've seen so far. Lookups on the set use hashes, so the membership test (highlighted) will be much quicker.

d = []
seen = set()
for bact in c:
    if bact[3] not in seen: # membership test
        seen.add(bact[3])
        d.append(bact)

Answer 3

Use pandas. I assume you have better column names as well.

c = [['470', '4189.0', 'asdfgw', 'fds'],
     ['470', '4189.0', 'qwer', 'fds'],
     ['470', '4189.0', 'qwer', 'dsfs fdv']]
import pandas as pd
df = pd.DataFrame(c, columns=['col_1', 'col_2', 'col_3', 'col_4'])
df.drop_duplicates('col_4', inplace=True)
print df

---

  col_1   col_2   col_3     col_4
0   470  4189.0  asdfgw       fds
2   470  4189.0    qwer  dsfs fdv