CODEWITHSUNDEEP
pythonlistelement
neonlights
neonlights

Reputation: 21

How do I get the program to print the number of times a number occurs in the list?

This is as much as I know how to do, not sure if I'm doing it correctly.

L = [4, 10, 4, 2, 9, 5, 4 ]

n = len(L)
element = ()


if element in L:
    print(element)

print("number occurs in list at the following position, if element not in list")
print("this number does not occur in the list")

How do I go about getting elements that appear more than once, to print as

4 occurs in L at the following positions:  [0, 2, 6]

Upvotes: 0

Views: 183

Answers (4)

Jon Clements
Jon Clements

Reputation: 142146

The compulsory defaultdict post:

from collections import defaultdict

el = [4, 10, 4, 2, 9, 5, 4 ]
dd = defaultdict(list)
for idx, val in enumerate(el):
    dd[val].append(idx)

for key, val in dd.iteritems():
    print '{} occurs in el at the following positions {}'.format(key, val)

#9 occurs in el at the following positions [4]
#10 occurs in el at the following positions [1]
#4 occurs in el at the following positions [0, 2, 6]
#2 occurs in el at the following positions [3]
#5 occurs in el at the following positions [5]

Then dd can just be used a normal dict... dd[4] or dd.get(99, "didn't appear")

Upvotes: 2

Rohit Jain
Rohit Jain

Reputation: 213233

You can use Counter to count distinct element in list, then use list comprehension to find the index of each element: -

>>> l = [4, 10, 4, 2, 9, 5, 4 ]
>>> from collections import Counter
>>> count = Counter(l)
>>> count
Counter({4: 3, 9: 1, 10: 1, 2: 1, 5: 1})

>>> lNew = [[(i,x) for i,x in enumerate(l) if x == cnt]  for cnt in count]
>>> 
>>> lNew[0]
[(4, 9)]
>>> lNew[1]
[(1, 10)]
>>> lNew[2]
[(0, 4), (2, 4), (6, 4)]
>>>

In fact you don't need Counter here. You can just get off with a Set

Create a Set of the list with the factory function : - set(l) and then you can do the same with it..

Upvotes: 0

Andy Hayden
Andy Hayden

Reputation: 375475

def print_repeated_elements_positions(L):
    for e in set(L): # only cover distinct elements
        if L.count(e) > 1: #only those which appear more than once
            print(e, "occurs at", [i for i, e2 in enumerate(L) if e2 == e])


L = [4, 10, 4, 2, 9, 5, 4]
print_repeated_elements_positions(L)
# prints: 4 occurs at [0, 2, 6]

Upvotes: 0

Tim Pietzcker
Tim Pietzcker

Reputation: 336148

You could use a list comprehension:

>>> L = [4, 10, 4, 2, 9, 5, 4]
>>> [i for i,x in enumerate(L) if x==4]
[0, 2, 6]

enumerate(L) gives you an iterator over L that yields a tuple (index, value) for each element of L. So what I'm doing here is take each index (i) if the value (x) equals 4, and construct a list from them. No need to look at the length of the list.

Upvotes: 1

Related Questions