I am running python 2.7.2 on a mac.
I have a simple dictionary:
dictionary= {a,b,c,a,a,b,b,b,b,c,a,w,w,p,r}
I want it to be printed and have the output like this:
Dictionary in alphabetical order:
a 4
b 5
c 2
p 1
r 1
w 2
But what I'm getting is something like this...
a 1
a 1
a 1
a 1
b 1
.
.
.
w 1
This is the code I am using.
new_dict = []
for word in dictionary.keys():
value = dictionary[word]
string_val = str(value)
new_dict.append(word + ": " + string_val)
sorted_dictionary = sorted(new_dict)
for entry in sorted_dictionary:
print entry
Can you please tell me where is the mistake? (By the way, I'm not a programmer but a linguist, so please go easy on me.)
What you're using is not a dictionary, it's a set ! :)
And sets doesn't allow duplicates.
What you probably need is not dictionaries, but lists.
A little explanation
Dictionaries have keys, and each unique keys have their own values:
my_dict = {1:'a', 2:'b', 3:'c'}
You retrieve values by using the keys:
>>> my_dict [1]
'a'
On the other hand, a list doesn't have keys.
my_list = ['a','b','c']
And you retrieve the values using their index:
>>> my_list[1]
'b'
Keep in mind that indices starts counting from zero, not 1.
Solving The Problem
Now, for your problem. First, store the characters as a list:
l = ['a', 'b', 'c', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'a', 'w', 'w', 'p', 'r']
Next, we'll need to know what items are in this list:
items = []
for item in l:
if item not in items:
items.append(item)
This is pretty much equal to items = set(l)
(the only difference is that this is a list). But just to make things clear, hope you understand what the code does.
Here is the content of items
:
>>> items
['a', 'b', 'c', 'w', 'p', 'r']
With that done, we will use lst.count()
method to see the number of a char's occurence in your list, and the built-in function sorted()
to sort the items:
for item in sorted(items): #iterates through the sorted items.
print item, l.count(item)
Result:
a 4
b 5
c 2
w 2
p 1
r 1
Hope this helps!!
Let's start with the obvious, this:
dictionary= {a,b,c,a,a,b,b,b,b,c,a,w,w,p,r}
is not a dictionary. It is a set
, and set
s do not preserve duplicates. You probably meant to declare that as a list
or a tuple
.
Now, onto the meat of your problem: you need to implement something to count the items of your collection. Your implementation doesn't really do that. You could roll your own, but really you should use a Counter
:
my_list = ['a','b','c','a','a','b','b','b','b','c','a','w','w','p','r']
from collections import Counter
c = Counter(my_list)
c
Out[19]: Counter({'b': 5, 'a': 4, 'c': 2, 'w': 2, 'p': 1, 'r': 1})
Now on to your next problem: dictionaries (of all types, including Counter
objects) do not preserve key order. You need to call sorted
on the dict's items()
, which is a list of tuples, then iterate over that to do your printing.
for k,v in sorted(c.items()):
print('{}: {}'.format(k,v))
a: 4
b: 5
c: 2
p: 1
r: 1
w: 2
dictionary is something like this {key1:content1, key2:content2, ...}
key in a dictionary is unique. then a = {1,2,3,4,5,5,4,5,6}
is the set, when you print this out, you will notice that
print a
set([1,2,3,4,5,6])
duplicates are eliminated.
In your case, a better data structure you can use is a list which can hold multiple duplicates inside.
if you want to count the element number inside, a better option is collections.Counter , for instance:
import collections as c
cnt = c.Counter()
dict= ['a','b','c','a','a','b','b','b','b','c','a','w','w','p','r']
for item in dict:
cnt[item]+=1
print cnt
the results would be:
Counter({'b': 5, 'a': 4, 'c': 2, 'w': 2, 'p': 1, 'r': 1})
as you notice, the results become a dictionary here.
so by using:
for key in cnt.keys():
print key, cnt[key]
you can access the key and content
a 4
c 2
b 5
p 1
r 1
w 2
you can achieve what you want by modifying this a little bit. hope this is helpful
for word in dictionary.keys():
, since here dictionary is not a dictionary type. If you like to write a code without using any builtin function, try this
input=['a','b','c','a','a','b','b','b','b','c','a','w','w','p','r']
dict={}
for x in input:
if x in dict.keys():
dict[x]=dict[x]+1
else:
dict[x]=1
for k in dict.keys():
print k, dict[k]
First, a dictionary is an unordered collection (ie, it has no guaranteed order of its keys).
Second, each dict key must be unique.
Though you could count the frequency of characters using a dict, there's a better the solution. The Counter
class in Python's collections
module is based on a dict and is specifically designed for a task like tallying frequency.
from collections import Counter
letters = ['a', 'b', 'c', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'a', 'w', 'w', 'p', 'r']
cnt = Counter(letters)
print cnt
The contents of the counter are now:
Counter({'b': 5, 'a': 4, 'c': 2, 'w': 2, 'p': 1, 'r': 1})
You can print these conveniently:
for char, freq in sorted(cnt.items()):
print char, freq
which gives:
a 4
b 5
c 2
p 1
r 1
w 2
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