列表python中元素（字符串）的计数

Question

genres_list = ['rock', 'pop', 'folk', 'dance', 'rusrap', 'ruspop', 'world', 'electronic', 'alternative', 'children', 'rnb', 'hip', 'jazz','postrock', 'latin', 'classical', 'metal', 'reggae', 'tatar','blues','pop', 'jazz']

# write function to count the number of specific genre in genres_list 

def find_genre(genre):
    count=0
    for count in genres_list:
        if count == genre:
            count=count+1
    return count

number=find_genre(pop)
print(number)

output:

TypeError: can only concatenate str (not "int") to str

Answer 1

Try this. list has a method to count the number occurences of an element.

genres_list = ['rock', 'pop', 'folk', 'dance', 'rusrap', 'ruspop', 'world', 'electronic', 'alternative', 'children', 'rnb', 'hip', 'jazz','postrock', 'latin', 'classical', 'metal', 'reggae', 'tatar','blues','pop', 'jazz']

print(genres_list.count('pop'))

---

output

2

list.count() complexity is O(n) .

You can write your own count function.

def find_genre(genre):
    count=0
    for _genre in genres_list:
        if _genre==genre:
            count+=1
    return count

print(find_genre('pop'))
#2

---

timeit analysis over a list of size 2 million. Results as of this writing (python 3.7, windows 10)

In [38]: timeit genres_list.count('pop') #My answer
26.6 ms ± 939 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [40]: timeit Counter(genres_list)['pop'] #Pitto's answer using collections.Counter
92.5 ms ± 751 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Now, custom count function everyone suggested including me.

def find_genre(genre):
    count=0
    for _genre in genres_list:
        if _genre==genre:
            count+=1
    return count

---

In [42]: timeit find_genre('pop')
63.9 ms ± 803 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Just for the sake of comparison ( _{Which I don't recommend using} ) I wrote some other functions to calculate count.

In [36]: timeit sum(list(map(lambda x:x=='pop',genres_list)))
334 ms ± 13.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [37]: timeit len(list(filter(lambda x:x=='pop',genres_list)))
188 ms ± 18.3 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [44]: timeit ' '.join(genres_list).count('pop')
41.5 ms ± 2.6 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Error occurred in your code because you use count for calculating the count of a genre and again you used count as a looping variable. In every iteration count becomes a str and we can't add str to a int type.

Answer 2

I would use Counter to achieve this result:

import collections

genres_list = ['rock', 'pop', 'folk', 'dance', 'rusrap', 'ruspop', 'world', 'electronic', 'alternative', 'children', 'rnb', 'hip', 'jazz','postrock', 'latin', 'classical', 'metal', 'reggae', 'tatar','blues','pop', 'jazz']

counted = collections.Counter(genres_list)
print(counted['rock'])

Output
1

Another possible solution, if you specifically want to fix your current approach:

def find_genre(genre):
    count=0
    for current_genre in genres_list:
        if current_genre == genre:
            count=count+1
    return count

number=find_genre('pop')
print(number)

Output
2

Your main issue was that naming the for loop variable in the same way that you named the counter variable was confusing you (and was behaving differently too).

As pointed out in comment from Johnny Mop:

This line for count in genres_list: makes count a string.

Answer 3

As Johnny Mopp already pointed out in the comments, for count in genres_list overwrites your count variable to be a string, so you could either do:

def find_genre(searched_genre):
    count=0
    for genre in genres_list:
        if searched_genre == genre:
            count = count+1
    return count

number = find_genre('pop')
print(number)

Or in a more pythonic way using list comprehension:

def find_genre(searched_genre):
    return len([genre for genre in genres_list if genre == searched_genre])

Or use a collection as @Pitto suggested (which might not be as intuitive when you first start programming)

Answer 4

First, you're using the same variable name in your iteration as the index in your list.

Second, you should use apostrophes in the argument pop like this 'pop' .

Hope this helps:

genres_list = ['rock', 'pop', 'folk', 'dance', 'rusrap', 'ruspop', 'world', 'electronic', 'alternative', 'children', 'rnb', 'hip', 'jazz','postrock', 'latin', 'classical', 'metal', 'reggae', 'tatar','blues','pop', 'jazz']

# write function to count the number of specific genre in genres_list
def find_genre(genre):
    num=0
    for count in genres_list:
        if count == genre:
            num=num+1
    return num

number=find_genre('pop')
print(number)