我有一個素數分解程式：
import sys
import math
def prime_factorize(n):
factors = []
number = math.fabs(n)
while number > 1:
factor = get_next_prime_factor(number)
factors.append(factor)
number /= factor
if n < -1:
factors[0] = -factors[0]
return tuple(factors)
def get_next_prime_factor(n):
if n % 2 == 0:
return 2
for x in range(3, int(math.ceil(math.sqrt(n)) 1), 2):
if n % x == 0:
return x
return int(n)
if __name__ == "__main__":
if len(sys.argv) != 2:
print("Usage: %s <integer>" % sys.argv[0])
exit()
try:
number = int(sys.argv[1])
except ValueError:
print("'%s' nie jest liczba calkowita!" % sys.argv[1])
else:
print ("%d =" %number, *prime_factorize(number), sep='*')
程式回傳 123456：
123456 =*2*2*2*2*2*2*3*643
我想以格式輸出：
123456 = 2^6*3*643
最好的解決方案是將元組轉換為字串？如何獲得這種輸出格式？
uj5u.com熱心網友回復：
我相信這就是你所追求的：
...
else:
tup = prime_factorize(number)
print("{} = {}".format(number, "*".join("{}^{}".format(x, tup.count(x)) if tup.count(x) > 1 else str(x) for x in sorted(set(tup)))))
輸出：
123456 = 2^6*3*643
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標籤：Python数组细绳元组