得到一个字典键的计数，其值大于python中的某个整数

Question

I have a dictionary. The keys are words the value is the number of times those words occur.

countDict = {'house': 2, 'who': 41, 'joey': 409, 'boy': 2, 'girl':2}

I'd like to find out how many elements occur with a value of more than 1, with a value of more than 20 and with a value of more than 50.

I found this code

a = sum(1 for i in countDict if countDict.values() >= 2)

but I get an error that I'm guessing means that values in dictionaries can't be processed as integers.

builtin.TypeError: unorderable types: dict_values() >= int()

I tried modifying the above code to make the dictionary value be an integer but that did not work either.

a = sum(1 for i in countDict if int(countDict.values()) >= 2)

builtins.TypeError: int() argument must be a string or a number, not 'dict_values'

Any suggestions?

Answer 1

You need this:

>>> countDict = {'house': 2, 'who': 41, 'joey': 409, 'boy': 2, 'girl':2}

>>> sum(1 for i in countDict.values() if i >= 2)
5

values() returns a list of all the values available in a given dictionary which means you can't convert the list to integer.

Answer 2

You could use collections.Counter and a "classification function" to get the result in one-pass:

def classify(val):
    res = []
    if val > 1:
        res.append('> 1')
    if val > 20:
        res.append('> 20')
    if val > 50:
        res.append('> 50')
    return res

from collections import Counter

countDict = {'house': 2, 'who': 41, 'joey': 409, 'boy': 2, 'girl':2}
Counter(classification for val in countDict.values() for classification in classify(val))
# Counter({'> 1': 5, '> 20': 2, '> 50': 1})

Of course you can alter the return values or thresholds in case you want a different result.

---

But you were actually pretty close, you probably just mixed up the syntax - correct would be:

a = sum(1 for i in countDict.values() if i >= 2)

because you want to iterate over the values() and check the condition for each value.

What you got was an exception because the comparison between

>>> countDict.values()
dict_values([2, 409, 2, 41, 2])

and an integer like 2 doesn't make any sense.

Answer 3

countDict.items() gives you key-value pairs in countDict so you can write:

>>> countDict = {'house': 2, 'who': 41, 'joey': 409, 'boy': 2, 'girl':2}
>>> [word for word, occurrences in countDict.items() if occurrences >= 20]
['who', 'joey']

If you just want the number of words, use len :

>>> countDict = {'house': 2, 'who': 41, 'joey': 409, 'boy': 2, 'girl':2}
>>> wordlist = [word for word, occurrences in countDict.items() if occurrences >= 20]
>>> len(wordlist)
2

Note that Python variables use lowercase and underscores rather than camelcase: count_dict rather than countDict . It's worth using this convention to avoid confusing yourself and others. See PEP8 for more details.

Answer 4

You're quite close. You just need to remember that i is accessible via the if statement. I added all three examples to get you started. Additionally, values creates a list of all values in the dictionary, which is not what you want, you instead want the currently evaluated value, i .

moreThan2 = sum(1 for i in countDict if countDict[i] >= 2)
moreThan20 = sum(1 for i in countDict if countDict[i] >= 20)
moreThan50 = sum(1 for i in countDict if countDict[i] >= 50)

Answer 5

Try using .items() before applying your discrimination logic.

in -

for key, value in countDict.items():
  if value == 2: #edit for use
     print key

out -

house
boy
girl