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>>> l = ["a","b","b"] >>> l.count("a") 1 >>> l.count("b") 2 Counting the occurrences of all items in a list is also known as "tallying" a list, or creating a tally counter. Counting all items with count() To count the occurrences of items in l one can simply use a list comprehension and the count() method [[x,l.count(x)] for x in set(l)]
str.count(a) is the best solution to count a single character in a string. But if you need to count more characters you would have to read the whole string as many times as characters you want to count. A better approach for this job would be:
All namespace modification in Python is a statement, for simplicity and consistency. That's one of the design decisions. And because integers are immutable, the only way to 'change' a variable is by reassigning it. Fortunately we have wonderful tools for the use-cases of ++ and --in other languages, like enumerate() and itertools.count().
item_count = 0 for item in list: item_count += 1 return item_count count([1,2,3,4,5]) (The list object must be iterable, implied by the for..in stanza.) The lesson here for new programmers is: You can’t get the number of items in a list without counting them at some point. The question becomes: when is a good time to count them?
The python cProfile and pstats modules offer great support for measuring time elapsed in certain functions without having to add any code around the existing functions. For example if you have a python script timeFunctions.py: import time def hello(): print "Hello :)" time.sleep(0.1) def thankyou(): print "Thank you!"
Python Count up & Down loop. Ask Question Asked 9 years ago. Modified 3 years, 9 months ago. Viewed 145k ...
I have a list of Booleans: [True, True, False, False, False, True] and I am looking for a way to count the number of True in the list (so in the example above, I want the return to be 3.) I have...
The existing solutions based on findall are fine for non-overlapping matches (and no doubt optimal except maybe for HUGE number of matches), although alternatives such as sum(1 for m in re.finditer(thepattern, thestring)) (to avoid ever materializing the list when all you care about is the count) are also quite possible.
You could count a single column by. df.A.count() #or df['A'].count() both evaluate to 5. The cool thing (or one of many w.r.t. pandas) is that if you have NA values, count takes that into consideration. So if I did. df['A'][1::2] = np.NAN df.count() The result would be. A 3 B 5
The main difference between groupby.count and groupby.size is that count counts only non-NaN values while size returns the length (which includes NaN), if the column has NaN values. value_counts() is equivalent to groupby.count by default but can become equivalent to groupby.size if dropna=False, i.e. df['col'].value_counts(dropna=False).