Searching and Filtering List of Tuples in Python

Question

I have a list of tuples, I'm trying to remove duplicates based on minimum value :

a_list = [("1","111","15"),("2","111","10"),("3","111","5"),("4","112","40"),("5","112","10")]

Output : 
id id_client value2
1   111         15
2   111         10
3   111         5
4   112         40
5   112         10

Required Output
id id_client value2
3   111         5
5   112         10

I tried everything but couldn't get it.

Show what you tried and what went wrong. And I guess your first "Output" is actually "Input"? — Michael Butscher
– Michael Butscher, Commented Oct 14, 2019 at 13:01
I answered a question like this earlier today: stackoverflow.com/questions/58369762/… (Except, instead of appending to the list, just keep hold onto the minimum value.) — Mateen Ulhaq
– Mateen Ulhaq, Commented Oct 14, 2019 at 13:07

Amit Yadav · Accepted Answer · 2019-10-14 13:14:57Z

1

Try the following code:

# Input list
a_list = [("1","111","15"),("2","111","10"),("3","111","5"),("4","112","40"),("5","112","10")]

# Sort the list by the third value (index-2)
sorted_list = sorted(a_list, key=lambda x: int(x[2]))

# Track visited and repeated elements to only add the first tuple(x) with the smallest x[1]
visited = []

# New list to only append unique tuples
new_list = []

for i in sorted_list:
    if i[1] not in visited:
        new_list.append(i)
        visited.append(i[1])
print(new_list)

Output:

[('1', '111', '15'), ('4', '112', '40')]

answered Oct 14, 2019 at 13:14

Amit Yadav

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2 Comments

dxtr Over a year ago

Thank you so much , worked perfectly, Would you please explain this line sorted_list = sorted(a_list, key=lambda x: int(x[2])) ?

Amit Yadav Over a year ago

Sure. It sorts the list by the third element x[2] of the tuples inside. You can use key attribute of sorted function for such operations. Simply provide a function to the key attribute and it will do the rest. Check out this more detail here

Mehrdad Pedramfar · Accepted Answer · 2019-10-14 13:18:16Z

1

Try this:

from itertools import groupby

new_l = []
for k,v in groupby(list, lambda x: x[1]):
    new_l.append(min(filter(lambda x:x[1]==k, list), key=lambda x:int(x[2])))

new_l will be your output.

Note that do not use pre-defiend names like list as a variable name. those names are mean something in python.

answered Oct 14, 2019 at 13:18

Mehrdad Pedramfar

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1 Comment

dxtr Over a year ago

Thank you so much , worked perfectly, I changed list to a_list in the question above

Aivar Paalberg · Accepted Answer · 2019-10-14 13:46:04Z

0

Another possibility without importing, just because comprehension is fun:

lst = [("1","111","15"),("2","111","10"),("3","111","5"),("4","112","40"),("5","112","10")] 

[min((x for x in lst if x[1] == client), key=lambda x: int(x[2])) for client in {row[1] for row in lst}]

gives

 [('5', '112', '10'), ('3', '111', '5')]

edited Oct 14, 2019 at 13:46

answered Oct 14, 2019 at 13:39

Aivar Paalberg

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