在迭代时将 function 值附加到列表

Question

I am a total Python newbie so pardon me for this stupid question

The composition() function returns a certain value

However I only want the value to be <=10 , and I want 100 of them The code below calculates the times it took to simulate 100 composition() values to be of <=10

def find():
    i=1
    aaa=composition()
    while(aaa>10):
        aaa=composition()
        i=i+1        
    return i

Customers_num=[find() for i in range(100)]
Customers_num=np.array(Customers_num)
print(np.sum(Customers_num))

However, Suppose that the code above returns 150. And I also want to know all the values that were being simulated in 150 times of composition() What kind of code should I start with?

I am thinking about combining it with the if else method statement and appending the values to an empty list, but so far my code has been a total disaster

def find():
    i=1
    aaa=composition()
    bbb=[]
    if aaa<=10:
        bbb.appendd([aaa])
    else:
        bbb.append([aaa])
        aaa=composition()
        bbb.appendd([aaa])
        while(aaa>10):
            i=i+1
            if aaa>10:
                bbb.append([aaa])
            else:
                bbb.append([aaa])            
    return i,bbb

find()

Thanks in advance!

Answer 1

You could make a generator that will generate n of list of values from composition() and stop when n of them are <= 10. This list will include all the numbers, so you have all the intermediate values, and the length of this list will be how "long" it took to generate it. For example (with a fake random composition() function:

from random import randint

def composition():
    return randint(0, 20)


def getN(n):
    while n > 0:
        c = composition()
        if c < 10:
            n -= 1
        yield c

values = list(getN(10)) # get 10 values less than 10
# [2, 0, 11, 15, 12, 8, 16, 16, 2, 8, 10, 3, 14, 2, 9, 18, 6, 11, 1]

time = len(values)
# 19