如何在python的二维数组中找到多少个值可以整除为某个值

Question

The following code generate random number of 2d arrays and I want to print how many values in each pair are divisible to 3. For example assume we have an array [[2, 10], [1, 6], [4, 8]]. So the first pair which is [2,10] has 3 ,6 and 9 which are totally 3 and second pair has 3 and 6 which are totally two and last pair[4,8] has just 1 divisible to 3 which is 6. Therefore, The final out put should print sum of total number of divisible values which is 3+2+1=6

a=random.randint(1, 10)

b = np.random.randint(1,10,(a,2))
b = [sorted(i) for i in b]   
c = np.array(b)            
   
counter = 0;   
  
for i in range(len(c)): 
    d=(c[i,0],c[i,1])
    if (i % 3 == 0):  
        counter = counter + 1
print(counter)

Answer 1

One way is to count how many integers in the interval are divisible by 3 by testing each one.

Another way, and this is much more efficient if your intervals are huge, is to use math.

Take the interval [2, 10] .

2 / 3 = 0.66; ceil(2 / 3) = 1 2 / 3 = 0.66; ceil(2 / 3) = 1 .

10 / 3 = 3.33; floor(10 / 3) = 3 10 / 3 = 3.33; floor(10 / 3) = 3 .

Now we need to count how many integers exist between 0.66 and 3.33, or count how many integers exist between 1 and 3. Hey, that sounds an awful lot like subtraction! (and then adding one)

Let's write this as a function

from math import floor, ceil
def numdiv(x, y, div):
    return floor(y / div) - ceil(x / div) + 1

So given a list of intervals, we can call it like so:

count = 0
intervals = [[2, 10], [1, 6], [4, 8]]
for interval in intervals:
    count += numdiv(interval[0], interval[1], 3)
print(count)

Or using a list comprehension and sum:

count = sum([numdiv(interval[0], interval[1], 3) for interval in intervals])

Answer 2

You can use sum() builtin for the task:

l = [[2, 10], [1, 6], [4, 8]]

print( sum(v % 3 == 0 for a, b in l for v in range(a, b+1)) )

Prints:

6

---

EDIT: To count number of perfect squares:

def is_square(n):
    return (n**.5).is_integer()

print( sum(is_square(v) for a, b in l for v in range(a, b+1)) )

Prints:

5

---

EDIT 2: To print info about each interval, just combine the two examples above. For example:

def is_square(n):
    return (n**.5).is_integer()

for a, b in l:
    print('Pair {},{}:'.format(a, b))
    print('Number of divisible 3: {}'.format(sum(v % 3 == 0 for v in range(a, b+1))))
    print('Number squares: {}'.format(sum(is_square(v) for v in range(a, b+1))))
    print()

Prints:

Pair 2,10:
Number of divisible 3: 3
Number squares: 2

Pair 1,6:
Number of divisible 3: 2
Number squares: 2

Pair 4,8:
Number of divisible 3: 1
Number squares: 1