Count how many times a dictionary value is found with more than one key

Question

I'm working in python. Is there a way to count how many times values in a dictionary are found with more than one key, and then return a count?

So if for example I had 50 values and I ran a script to do this, I would get a count that would look something like this:

1: 23  
2: 15  
3: 7  
4: 5  

The above would be telling me that 23 values appear in 1 key, 15 values appear in 2 keys, 7 values appear in 3 keys and 5 values appear in 4 keys.

Also, would this question change if there were multiple values per key in my dictionary?

Here is a sample of my dictionary (it's bacteria names):

{'0': ['Pyrobaculum'], '1': ['Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium'], '3': ['Thermoanaerobacter', 'Thermoanaerobacter'], '2': ['Helicobacter', 'Mycobacterium'], '5': ['Thermoanaerobacter', 'Thermoanaerobacter'], '4': ['Helicobacter'], '7': ['Syntrophomonas'], '6': ['Gelria'], '9': ['Campylobacter', 'Campylobacter'], '8': ['Syntrophomonas'], '10': ['Desulfitobacterium', 'Mycobacterium']}

So from this sample, there are 8 unique values, I the ideal feedback I would get be:

1:4
2:3
3:1

So 4 bacteria names are only in one key, 3 bacteria are found in two keys and 1 bacteria is found in three keys.

Answer 1

If I understand correctly, you want to count the counts of dictionary values. If the values are countable by collections.Counter , you just need to call Counter on the dictionaries values and then again on the first counter's values. Here is an example using a dictionary where the keys are range(100) and the values are random between 0 and 10:

from collections import Counter
d = dict(enumerate([str(random.randint(0, 10)) for _ in range(100)]))
counter = Counter(d.values())
counts_counter = Counter(counter.values())

EDIT :

After the sample dictionary was added to the question, you need to do the first count in a slightly different way ( d is the dictionary in the question):

from collections import Counter
c = Counter()
for v in d.itervalues():
    c.update(set(v))
Counter(c.values())

Answer 2

So unless I'm reading this wrong you want to know:

• For each of the values in the original dictionary, how many times does each different count of values occur?
• In essence what you want is the frequency of the values in the dictionary

I took a less elegant approach that the other answers, but have broken the problem down for you into individual steps:

d = {'0': ['Pyrobaculum'], '1': ['Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium'], '3': ['Thermoanaerobacter', 'Thermoanaerobacter'], '2': ['Helicobacter', 'Mycobacterium'], '5': ['Thermoanaerobacter', 'Thermoanaerobacter'], '4': ['Helicobacter'], '7': ['Syntrophomonas'], '6': ['Gelria'], '9': ['Campylobacter', 'Campylobacter'], '8': ['Syntrophomonas'], '10': ['Desulfitobacterium', 'Mycobacterium']}

# Iterate through and find out how many times each key occurs
vals = {}                       # A dictonary to store how often each value occurs.
for i in d.values():
  for j in set(i):              # Convert to a set to remove duplicates
    vals[j] = 1 + vals.get(j,0) # If we've seen this value iterate the count
                                # Otherwise we get the default of 0 and iterate it
print vals

# Iterate through each possible freqency and find how many values have that count.
counts = {}                     # A dictonary to store the final frequencies.
# We will iterate from 0 (which is a valid count) to the maximum count
for i in range(0,max(vals.values())+1):
    # Find all values that have the current frequency, count them
    #and add them to the frequency dictionary
    counts[i] = len([x for x in vals.values() if x == i])

for key in sorted(counts.keys()):
  if counts[key] > 0:
     print key,":",counts[key]

You can also test this code on codepad .

Answer 3

You could use a Counter

>>>from collections import Counter
>>>d = dict(((1, 1), (2, 1), (3, 1), (4, 2), (5, 2), (6, 3), (7, 3)))
>>>d
{1: 1, 2: 1, 3: 1, 4: 2, 5: 2, 6: 3, 7: 3}
>>>Counter(d.values())
Counter({1: 3, 2: 2, 3: 2})