CODEWITHSUNDEEP
function
Alan Wakke
Alan Wakke

Reputation: 87

Finding the min

I'm a beginner in python and having trouble with this function. I'm trying to count the values in list xs that are greater than v as well as the duplicates. I'm getting an error for line 3:

NameError: name 'j' is not defined

Code below that I have:

def count_greater(xs,v):
    count_list = 0
    for i in (xs) and j in (v):
        if i > j:
            count_list += 1 
    else:
        count_list += 0
    return count_list 
count_greater([12,0,20,34,0,20],3)

How should I approach this? I am unsure on how to complete the rest of the logic.

Upvotes: 0

Views: 66

Answers (6)

Moinuddin Quadri
Moinuddin Quadri

Reputation: 48077

Your for loop is syntactically incorrect. You need to iterate over the just xslist and check with the value of v. For example to get number greater than and equal to v in your xs:

def count_get_greater_item(xs,v):
    greater_than_v = []
    for item in xs:
        if item >= v:
            greater_than_v.append(item)
    return greater_than_v 

count_above([22,0,30,44,0,30],5)

However, for finding the duplicates (my understanding is the number occurring in the list more than once), you may use collections.Counter() as:

import collections
duplicate_list = []
for item, count in collections.Counter(x).items():
    if count > v:
         duplicate_list.append(item)

OR, in one line via list comprehension as:

duplicate_list = [item for item, count in collections.Counter(xs).items() if count > v]

where xs and v are holding the same value as is mentioned in question

Upvotes: 0

techydesigner
techydesigner

Reputation: 1691

I think this is what you are after: 

def count_greater(xs,v):
    count = 0
    for i in xs:
        if i >= v:
            count += 1
    return count
count_greater([12,0,20,34,0,20],3)

It returns the number of times that a value in the list is equal to or greater than the second argument (3 in the example). The value returned in this case is 4.

Upvotes: 0

Apollo2020
Apollo2020

Reputation: 519

You can use a list comprehension to get conditional occurrence counts and sets to identify the (total) number of duplicates:

vals = [22,0,30,44,0,30]
ceil = 5

over_ceil_count = len([v for v in vals if v > ceil])

dup_count = len(vals) - len(set(vals))

The set automatically removes duplicates, so the difference between the length of the set and the length of the original list tells you how many duplicate values you started with.

Upvotes: 0

Prune
Prune

Reputation: 77847

Use the built-in capabilities of the language:

def count_greater(xs,v):
    return sum([item >= v for item in xs])

print (count_greater([12,0,20,34,0,20],3))

Output is 4, as expected.

In your code, note that count += 0 does absolutely nothing; remove that and the else. As you suggested, just use v; there's no need to iterate j over a construct that you know is exactly one item long.

Upvotes: 0

RecencyEffect
RecencyEffect

Reputation: 756

v is not a list, but a single number, so you do not need to iterate over it.

So this would be sufficient:

for i in (xs):
    if i > v:
        count_list += 1

There are alternative approaches you could take. For instance, use filter(function, list), like so:

list_greater = filter(lambda d: d < v, xs)
count_greater = len(list_greater)

This approach has the advantage that you can work with the filtered numbers, too, not just count them.

Upvotes: 1

user7179690
user7179690

Reputation: 1123

as I understand the working code is it count the values in the list more than v 

def count_above(xs,v):
    count_list = 0
    for i in xs: # for each i in the list which is xs
        if i > v:
            count_list += 1 
        else: # take care of else space should be under if dircitly
            count_list += 0
    return count_list 

count_above([22,0,30,44,0,30],5)

you should take care of your spaces and you loop on a list for each i n the list as i mention in code above

Upvotes: 0

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