Как найти количество элементов в объекте Python? Подсчитываем число элементов в списке, массиве, кортеже
В этой статье мы рассмотрим, как определить количество элементов в объекте Python и при необходимости подсчитать их сумму. Также увидим, как подсчитать количество вхождений конкретного элемента.
Итак, представим, что у нас есть следующий массив:
По условию задачи мы хотим определить, сколько элементов в данном массиве, и какова сумма всех этих элементов.
В первую очередь, вспомним, что в языке программирования Python существует специальная функция, возвращающая длину списка, массива, последовательности и так далее — это len(x) , где x — наша последовательность.
Если разобраться, длина последовательности из чисел — это одновременно и количество самих цифр, поэтому мы можем решить поставленную задачу следующим образом:
print(len(array)) 6 Press any key to continue . . .А для подсчёта суммы можем занести перечисление массива Python в цикл:
array = [6,2,7,4,8,1] sum = 0 for i in range(len(array)): sum = array[i] print(sum)В принципе, вопрос решён. Но, по правде говоря, перебор целочисленного массива с помощью цикла для получения суммы элементов массива — это, всё же, костыль)). Дело в том, что в Python существует встроенная функция sum() . Она вернёт нам сумму без лишних телодвижений.
def main(): array = [1,6,3,8,4,9,25,2] print(sum(array)) if name == 'main': main() 58 Press any key to continue . . .Python: количество вхождений конкретного элемента
Бывает, нам надо подсчитать число вхождений определённых элементов в списке и вернуть найденное значение. Для этого в Python есть метод count() . Вот его синтаксис:
Метод принимает аргумент x, значение которого нас интересует. И возвращает число вхождений интересующего элемента в список:
# объявляем список website_list = ['otus.ru','includehelp.com', 'yandex.by', 'otus.ru'] # подсчитываем вхождения 'otus.ru' count = website_list.count('otus.ru') print('otus.ru found',count,'times.') # подсчитываем вхождения 'yandex.by' count = website_list.count('yandex.by') print('yandex.by found',count,'times.')otus.ru found 2 times. yandex.by found 1 times.Также этот метод успешно работает и с кортежами:
# объявляем кортеж sample_tuple = ((1,3), (2,4), (4,6)) # условные вхождения (1,2) count = sample_tuple.count((1,2)) print('(1,2) found',count,'times.') # условные вхождения (1,3) count = sample_tuple.count((1,3)) print('(1,3) found',count,'times.')(1,2) found 0 times. (1,3) found 1 times.Вот и всё, теперь вы знаете, как подсчитывать количество элементов в списке, массиве, кортеже в Python.
Count the number of occurrences of a character in a string
How do I count the number of occurrences of a character in a string? e.g. 'a' appears in 'Mary had a little lamb' 4 times.
You might find the simplest way to code it but at the end, time complexity remains the same ,whether we use loops or built in count() .
26 Answers 26
str.count(sub[, start[, end]])
Return the number of non-overlapping occurrences of substring sub in the range [start, end] . Optional arguments start and end are interpreted as in slice notation.
>>> sentence = 'Mary had a little lamb' >>> sentence.count('a') 4@RufusVS Just to mention, that doesn't work for all writing systems. For a more thorough approach, see Veedrac's answer on "How do I do a case-insensitive string comparison?"
>>> 'Mary had a little lamb'.count('a') 4To get the counts of all letters, use collections.Counter :
>>> from collections import Counter >>> counter = Counter("Mary had a little lamb") >>> counter['a'] 4If you want the counts for a lot of the letters in a given string, Counter provides them all in a more succinct form. If you want the count for one letter from a lot of different strings, Counter provides no benefit.
For this particular instance, counting characters, I would prefer collections.counter. For finding instances of a specific substring, I would use a regular expression or the str.count() method. I haven't tested, but there may be a performance difference due to a slight overhead in counting all characters and appending to a dictionary rather than counting occurrences of a single substring. I would suggest writing a script to generate a very long file to search and then timing execution of each method.
The advantage when used frequently is that Counter calculates all the counts ONE TIME, which is almost as fast as doing mystring.count('a') one time. Thus, if you do this 20 times, you are saving maybe 10 times the computation time. Counter also can tell you if an item is in the string: for example, if 'a' in counter:
Regular expressions maybe?
import re my_string = "Mary had a little lamb" len(re.findall("a", my_string))A fine idea, but overkill in this case. The string method 'count' does the same thing with the added bonus of being immediately obvious about what it is doing.
This should be downvoted because it is the least efficient way possible to count characters in a string. If the goal is simply to count characters, as the question indicates, it would be hard to find a worse way to do the job. In terms of memory and processor overhead, this solution is definitely to be avoided. No one will ever "need" to use this method to find the count of characters in a string.
str.count(sub[, start[, end]])
Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation.
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:
from collections import defaultdict text = 'Mary had a little lamb' chars = defaultdict(int) for char in text: chars[char] += 1So you'll have a dict that returns the number of occurrences of every letter in the string and 0 if it isn't present.
For a case insensitive counter you could override the mutator and accessor methods by subclassing defaultdict (base class' ones are read-only):
class CICounter(defaultdict): def __getitem__(self, k): return super().__getitem__(k.lower()) def __setitem__(self, k, v): super().__setitem__(k.lower(), v) chars = CICounter(int) for char in text: chars[char] += 1 >>>chars['a'] 4 >>>chars['M'] 2 >>>chars['x'] 0@merv Not really. Counter is a more bloated pure Python class and defaultdict 's __missing__ is written in C. For a simple task like this ( int is also implemented in C) this approach is sligthly faster.
This easy and straightforward function might help:
def check_freq(x): freq = <> for c in set(x): freq[c] = x.count(c) return freq check_freq("abbabcbdbabdbdbabababcbcbab")If a comprehension is desired:
Regular expressions are very useful if you want case-insensitivity (and of course all the power of regex).
my_string = "Mary had a little lamb" # simplest solution, using count, is case-sensitive my_string.count("m") # yields 1 import re # case-sensitive with regex len(re.findall("m", my_string)) # three ways to get case insensitivity - all yield 2 len(re.findall("(?i)m", my_string)) len(re.findall("m|M", my_string)) len(re.findall(re.compile("m",re.IGNORECASE), my_string))Be aware that the regex version takes on the order of ten times as long to run, which will likely be an issue only if my_string is tremendously long, or the code is inside a deep loop.
Regex is overkill if you are just trying to fix case sensitivity. my_sting.lower().count('m') is more performant, more clear, and more succinct.
I don't know about 'simplest' but simple comprehension could do:
>>> my_string = "Mary had a little lamb" >>> sum(char == 'a' for char in my_string) 4Taking advantage of built-in sum, generator comprehension and fact that bool is subclass of integer: how may times character is equal to 'a'.
a = 'have a nice day' symbol = 'abcdefghijklmnopqrstuvwxyz' for key in symbol: print(key, a.count(key))I am a fan of the pandas library, in particular the value_counts() method. You could use it to count the occurrence of each character in your string:
>>> import pandas as pd >>> phrase = "I love the pandas library and its `value_counts()` method" >>> pd.Series(list(phrase)).value_counts() 8 a 5 e 4 t 4 o 3 n 3 s 3 d 3 l 3 u 2 i 2 r 2 v 2 ` 2 h 2 p 1 b 1 I 1 m 1 ( 1 y 1 _ 1 ) 1 c 1 dtype: int64An alternative way to get all the character counts without using Counter() , count and regex
counts_dict = <> for c in list(sentence): if c not in counts_dict: counts_dict[c] = 0 counts_dict[c] += 1 for key, value in counts_dict.items(): print(key, value)count is definitely the most concise and efficient way of counting the occurrence of a character in a string but I tried to come up with a solution using lambda , something like this:
sentence = 'Mary had a little lamb' sum(map(lambda x : 1 if 'a' in x else 0, sentence))Also, there is one more advantage to this is if the sentence is a list of sub-strings containing same characters as above, then also this gives the correct result because of the use of in . Have a look:
sentence = ['M', 'ar', 'y', 'had', 'a', 'little', 'l', 'am', 'b'] sum(map(lambda x : 1 if 'a' in x else 0, sentence))But of course this will work only when checking occurrence of single character such as 'a' in this particular case.
a = "I walked today," c=['d','e','f'] count=0 for i in a: if str(i) in c: count+=1 print(count)Hi @GinoMempin don't think the intention here is to differentiate however you may declare two other variables and compare i to 'd','e' and 'f' separately if you wish to do so.
I know the ask is to count a particular letter. I am writing here generic code without using any method.
sentence1 =" Mary had a little lamb" count = <> for i in sentence1: if i in count: count[i.lower()] = count[i.lower()] + 1 else: count[i.lower()] = 1 print(count)Now if you want any particular letter frequency, you can print like below.
the easiest way is to code in one line:
'Mary had a little lamb'.count("a")but if you want can use this too:
sentence ='Mary had a little lamb' count=0; for letter in sentence : if letter=="a": count+=1 print (count)Twenty-two answers. The top answer has more upvotes than I’ve received in total over eight years on Stack Overflow. Why do you prefer this solution? What is it contributing that the existing answers are missing?
Now, I see they mentioned it! It is ok I just wanted to show my code too. I think no one mentioned it
Ther are two ways to achieve this:
1) With built-in function count()
sentence = 'Mary had a little lamb' print(sentence.count('a'))`2) Without using a function
sentence = 'Mary had a little lamb' count = 0 for i in sentence: if i == "a": count = count + 1 print(count)To find the occurrence of characters in a sentence you may use the below code
Firstly, I have taken out the unique characters from the sentence and then I counted the occurrence of each character in the sentence these includes the occurrence of blank space too.
ab = set("Mary had a little lamb") test_str = "Mary had a little lamb" for i in ab: counter = test_str.count(i) if i == ' ': i = 'Space' print(counter, i)Output of the above code is below.
1 : r , 1 : h , 1 : e , 1 : M , 4 : a , 1 : b , 1 : d , 2 : t , 3 : l , 1 : i , 4 : Space , 1 : y , 1 : m ,Downvote. Duplicate of stackoverflow.com/a/49385352/11154841, it only makes the unneeded way over ''.join() .
@-questionto42 may you please elaborate the duplication meaning here stated by you but I agreed with your point over use of join and i am removing the same. Thank you!
The other answer was first and also has a set() of a string in it that it loops over to check the count() of each letter in the set. It adds the results to keys of a dictionary and prints the dictionary afterwards instead of this answer that prints the counts directly during the loop (how the results are printed is not the main idea anyway). Therefore the duplicate.
@questionto42 The concept can be used by anyone and the same idea may come into different mind but when you say the answer is duplicate I say it is not as I have gone through that stackoverflow that you mention in the comment and there is a difference in that answer and mine there space character is not been calculated and in my answer it is been calculated so its not a duplicate as duplicate means each and everything should be exactly same. please make a note on this and if you find this explanation correct than you may remove your down vote.
I had a look at it. The other solution gives you a dictionary with a ' ' as the key and the number of ' ' as the value. You can rename a key of a dictionary as you like using pop , in this case ' ' to space , this does not add value. I still withdraw my downvote since someone might want to have a solution without a dictionary. Yet, this is a duplicate, the main trick is just the set() and the count() , which you repeat.