如何对字典中的项目进行排序并显示最大的键和值？

Question

I wrote a program that reads 10 numbers from the input and at the end prints the number that has the largest number of divisors of the first number along with the number of divisors of the first number in the output. If several have this mode, print the largest of them.

The input I give to the program:

1854
9875
3567
2568
2984
5428
6487
7982
3485
2157

import operator
number_list= []
divisor_prime_list = []
dwq = {}
sd = []
for i in range(10):
    n = int(input())
    number_list.append(n)
    i = 2
    a = {}
    factors = []
    while i*i<=n :
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
         
    if n > 1:
        factors.append(n)
        fu = list(set(factors))
        divisor_prime_list.append(len(fu))
for d in number_list:
    for iq in divisor_prime_list:
        dwq[d] = iq
        divisor_prime_list.remove(iq)
        break
b = reversed(sorted(dwq.items(),key=operator.itemgetter(0)))
for k,v in b:
    print(k,v)
    break

But I had trouble displaying the output

My code output is incorrect:

9875 2

And the correct output:

7982 3

Please help me edit the code so that it shows me the correct output

Answer 1

I would recommend writing little testable building blocks for your problem, in the form of functions.

For example:

from math import gcd
from itertools import combinations

def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return factors

def factor_count(x):
    primes = prime_factors(x)
    return len([t for n in range(1, len(primes) + 1)
                for t in set(combinations(primes, n))])

Application:

a = 1854
b = [9875, 3567, 2568, 2984, 5428, 6487, 7982, 3485, 2157]

ndiv = [factor_count(gcd(a, x)) for x in b]
max([(n, x) for n, x in zip(ndiv, b)])

# out:
(3, 2568)

If you must take user input etc., then:

values = [int(input()) for _ in range(10)]
a, b = values[0], values[1:]

# and then use the above

---

Supplement:

# while not immediately needed in your question, you may like:
from functools import reduce

def prod(a):
    return reduce(lambda x, y: x * y, a)

def factors(x):
    primes = prime_factors(x)
    for n in range(1, len(primes) + 1):
        for t in set(combinations(primes, n)):
            yield prod(t)

Example:

>>> prime_factors(20)
[2, 2, 5]

>>> sorted(factors(20))
[2, 4, 5, 10, 20]

>>> factor_count(20)
5

# list of each pair under consideration, along with common divisors:
>>> [(a, x, sorted(factors(gcd(a, x)))) for x in b]
[(1854, 9875, []),
 (1854, 3567, [3]),
 (1854, 2568, [2, 3, 6]),
 (1854, 2984, [2]),
 (1854, 5428, [2]),
 (1854, 6487, []),
 (1854, 7982, [2]),
 (1854, 3485, []),
 (1854, 2157, [3])]

Answer 2

Here is your code fixed:

number_list = []
divisor_prime_list = []
dwq = []
sd = []
for j in range(10):
    n = int(input())
    number_list.append(n)
    i = 2
    a = {}
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)

    if n > 1:
        factors.append(n)

    fu = set(factors)
    divisor_prime_list.append(len(fu))

for d, p in zip(number_list, divisor_prime_list):
    dwq.append((p, d))

b = max(dwq)
print(b[1], b[0])

The loop variable i of the outer for was the same as the one you used to find the primes.

You forgot to dedent at fu = list(set(factors))

I haven't been able to understand your intentions after for d in number_list , so I rewrote that heavily. All you need to do is collect the pairs of numbers and their divisor count. If you put the divisor count first, all you need to do is find the max of the pairs. Remember that tuples are sorted (by default) by their first element, then by their second, and so on.