CODEWITHSUNDEEP
python
user3344650
user3344650

Reputation: 23

How to remove duplicate from this?

a= {'bana':3, 'hello':2, 'clue':7, 'an':1,'sad':3}
b=list(a.keys())
c=list(a.values())
sorted(c,reverse=True)
x=sorted(c, reverse=True)
print(x)
print(b)
d=[]
for i in x:
   for z in b:
       if a[z] == i:
           l= str(i) +z
           d.append(l)
b=sorted(d, reverse=True)
print(d)

hello, I have this code above. Whenever I print out d, I get:

['7clue', '3sad', '3bana', '3sad', '3bana', '2hello', '1an']

is there anyhow I can make it so that it is 

['7clue', '3sad', '3bana', '2hello', '1an']

so it doesn't have that repeated 2 entries? Thanks

Upvotes: 0

Views: 75

Answers (5)

Barraka
Barraka

Reputation: 52

Also, this should work:

a= {'bana':3, 'hello':2, 'clue':7, 'an':1,'sad':3}
b=list(a.keys())
x=sorted(list(a.values()), reverse=True)
d=[]
for i in range(len(x)):
    d.append(str(x[i])+b[i])
print(d)

Upvotes: 0

ssm
ssm

Reputation: 5373

How about:

In [10]: b = [str(v)+k  for k, v in zip(a.keys(), a.values())]

In [11]: b
Out[11]: ['3sad', '1an', '3bana', '2hello', '7clue']

In [12]: sorted(b, key = lambda m: int(m[0:1] ))
Out[12]: ['1an', '2hello', '3sad', '3bana', '7clue']

In [13]: sorted(b, key = lambda m: int(m[0:1] ), reverse=True)
Out[13]: ['7clue', '3sad', '3bana', '2hello', '1an']

?

Upvotes: 0

John La Rooy
John La Rooy

Reputation: 304137

This is the unique_everseen recipe from the itertools docs

from itertools import ifilterfalse, 
def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

L = ['7clue', '3sad', '3bana', '3sad', '3bana', '2hello', '1an']
res = list(unique_everseen(L))

If order doesn't matter, you can of course just use

L = list(set(L))

Upvotes: 0

Slater Victoroff
Slater Victoroff

Reputation: 21914

There are a couple ways of doing it, but I think this would be the most efficient so long as ordering doesn't matter:

>>> a = ['7clue', '3sad', '3bana', '3sad', '3bana', '2hello', '1an']
>>> list(set(a))
['1an', '2hello', '7clue', '3bana', '3sad']

If order does matter, then you can do this with a simple list comp:

>>> def remove_dups(a):
...     seen = set()
...     seen_add = seen.add
...     return [ x for x in a if x not in seen and not seen_add(x)]
... 
>>> remove_dups(['7clue', '3sad', '3bana', '3sad', '3bana', '2hello', '1an'])
['7clue', '3sad', '3bana', '2hello', '1an']

Upvotes: 1

hobenkr
hobenkr

Reputation: 1234

There may be a better way, but I've found using set() works well enough for most of what I do. For example, set(d). If you'd like the type to stay a list, you could use list(set(d)).

Upvotes: 0

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