So I have found the frequencies of numbers from a list and have created a list such as this [2:3 , 25:1, 22:4, 5:2, 5:2] What im trying to do after that is have a for loop detecting the maximum frequency (different numbers can have the same frequencies) and then printing the number and frequency that is the highest which may be more than one.
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7Do you have a `list` or a `dict`? Because `[2:3 , 25:1, 22:4, 5:2, 5:2]` is neither. – mata May 30 '12 at 07:39
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So, in `2:3`, which is the number and which is the frequency? Also, what is the significance of `5:2` appearing twice? – NPE May 30 '12 at 07:41
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So number is 2 and frequency is 3. Umm trying to show that different numbers can have the same max frequency – bluelantern May 30 '12 at 07:43
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Are you trying to do something like this? http://stackoverflow.com/questions/268272/getting-key-with-maximum-value-in-dictionary – acattle May 30 '12 at 07:44
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My guess is for 2:3 you get the frequency when you divide 2 by 3. @bluelantem some clarification would help. What's the datatype of a single list entry? – Nicola Coretti May 30 '12 at 07:46
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The Number is 2, the frequency is 3 – bluelantern May 30 '12 at 07:47
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1So it's a dictionary? I.e. {2:3 , 25:1, 22:4, 5:2}? Why does 5:2 appear twice? Is this a homework assignment? – acattle May 30 '12 at 07:58
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Yes my bad sorry a dictionary – bluelantern May 30 '12 at 08:17
1 Answers
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Depending on how your data is structured
>>> data = {2:3 , 25:1, 22:4, 5:2, 5:2}
>>> max(data, key = lambda x: data[x])
22
or
>>> data = [(2, 3), (25, 1), (22, 4), (5,2), (5,2)]
>>> max(data, key = lambda x: x[1])
(22, 4)
should do the trick.
[Edit]
>>> data = {2:3 , 25:4, 22:4, 5:2, 5:2}
>>> max_key = max(data.values())
>>> print [i for i in data if data[i] == max_key]
[22, 25]
luke14free
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Lets say you have data = {2:4, 5:4, 3:2, 22:6} how would you display 2 and 5 – bluelantern May 30 '12 at 08:24
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1Edited, this will take into account multiple values. You are using a bad data structure though. I strongly suggest you to take a look at http://docs.python.org/dev/library/collections.html#collections.Counter – luke14free May 30 '12 at 08:29