问题
I'm truly a beginner at python so I apologise for the lack of knowledge, but the reason I'm asking is that reading the Python manual and tutorial (http://docs.python.org/2.7/tutorial) I'm not unable to totally grasp how loops work. I've written some simple programs so I think I get the basics but for whatever reason this program that is meant to list all primes less than or equal to n is not working:
n = int(raw_input("What number should I go up to? "))
p = 2
while p <= n:
for i in range(2, p):
if p%i == 0:
p=p+1
print "%s" % p,
p=p+1
print "Done"
The output when I enter 100 for example is:
2 3 5 7 11 13 17 19 23 27 29 31 35 37 41 43 47 53 59 61 67 71 73 79 83 87 89 95 97 101 Done
Which looks almost right but for some reason contains 27, 35, 95, which are composite of course. I've been trying to pick apart the way my loop works but I just don't see where it skips checking for divisibility suddenly. I figured that if someone had a look they could explain to me what about the syntax is causing this. Thanks a bunch!
回答1:
I would actually restructure the program to look like this:
for p in range(2, n+1):
for i in range(2, p):
if p % i == 0:
break
else:
print p,
print 'Done'
This is perhaps a more idiomatic solution (using a for loop instead of a while loop), and works perfectly.
The outer for loop iterates through all the numbers from 2 to n.
The inner one iterates to all numbers from 2 to p. If it reaches a number that divides evenly into p, then it breaks out of the inner loop.
The else block executes every time the for loop isn't broken out of (printing the prime numbers).
Then the program prints 'Done' after it finishes.
As a side note, you only need to iterate through 2 to the square root of p, since each factor has a pair. If you don't get a match there won't be any other factors after the square root, and the number will be prime.
回答2:
Your code has two loops, one inside another. It should help you figure out the code if you replace the inner loop with a function. Then make sure the function is correct and can stand on its own (separate from the outer loop).
Here is my rewrite of your original code. This rewrite works perfectly.
def is_prime(n):
i = 2
while i < n:
if n%i == 0:
return False
i += 1
return True
n = int(raw_input("What number should I go up to? "))
p = 2
while p <= n:
if is_prime(p):
print p,
p=p+1
print "Done"
Note that is_prime() doesn't touch the loop index of the outer loop. It is a stand-alone pure function. Incrementing p inside the inner loop was the problem, and this decomposed version doesn't have the problem.
Now we can easily rewrite using for loops and I think the code gets improved:
def is_prime(n):
for i in range(2, n):
if n%i == 0:
return False
return True
n = int(raw_input("What number should I go up to? "))
for p in range(2, n+1):
if is_prime(p):
print p,
print "Done"
Note that in Python, range() never includes the upper bound that you pass in. So the inner loop, which checks for < n, we can simply call range(2, n) but for the outer loop, where we want <= n, we need to add one to n so that n will be included: range(2, n+1)
Python has some built-in stuff that is fun. You don't need to learn all these tricks right away, but here is another way you can write is_prime():
def is_prime(n):
return not any(n%i == 0 for i in range(2, n))
This works just like the for loop version of is_prime(). It sets i to values from range(2, n) and checks each one, and if a test ever fails it stops checking and returns. If it checks n against every number in the range and not any of them divide n evenly, then the number is prime.
Again, you don't need to learn all these tricks right away, but I think they are kind of fun when you do learn them.
回答3:
This should work and is bit more optimized
import math
for i in range(2, 99):
is_prime = True
for j in range(2, int(math.sqrt(i)+1)):
if i % j == 0:
is_prime = False
if is_prime:
print(i)
回答4:
Please compare your snippet with the one pasted below and you will notice where you were wrong.
n = int(raw_input("What number should I go up to? "))
p = 2
while p <= n:
is_prime=True
for i in range(2, p):
if p%i == 0:
is_prime=False
break;
if is_prime==True:
print "%d is a Prime Number\n" % p
p=p+1
回答5:
you do not re-start the i loop after you find a non-prime
p = i = 2
while p <= n:
i = 2
while i < p:
if p%i == 0:
p += 1
i = 1
i += 1
print p,
p += 1
print "Done"
A while loop executes the body, and then checks if the condition at the top is True, if it is true, it does the body again. A for loop executes the body once for each item in the iterator.
回答6:
def is_prime(n):
if n>=2:
for i in range(2, n):
if n%i == 0:
return False
return True
else:
return False
To find PRIME NUMBER
回答7:
Let's do a couple more improvements.
- You know 2 is the only even prime number, so you add 2 in your list and start from 3 incrementing your number to be checked by 2.
- Once you are past the half-way point (see above
sqrtand*examples), you don't need to test for a prime number. - If you use a list to keep track of the prime numbers, all you need to do is to divide by those prime numbers.
I wrote my code and each of the above items would improve my code execution time by about 500%.
prime_list=[2]
def is_prime(a_num):
for i in prime_list:
div, rem = divmod(a_num, i)
if rem == 0:
return False
elif div < i:
break;
prime_list.append(a_num)
return True
回答8:
print('Enter a Number: ')
number=abs(int(input()))
my_List=[0,1]
def is_prime(n):
if n in my_List:
return True
elif n>=2:
for i in range(2, n):
if n%i == 0:
return False
return True
else:
return False
if is_prime(number):
print("%d is Prime!"%number)
else:
print(number,'is not prime')
回答9:
for i in range(2, p):
if p%i == 0:
p=p+1
print "%s" % p,
p=p+1
I am going to tell your error only,in line 3 you are incrimenting p but actually what you are missing is your i if your i in previous case is let say 13 then it will check your loop after 13 but it is leaving 2,3,5,7,11 so its an error .that is what happening in case of 27 your i before 27 is 13 and now it will check from 14.and I don't think u need an solution.
回答10:
def findprime(num):
count = 0
for i in range(1,num+1):
list1 = []
for ch in range(1,i+1):
if i%1==0 and i%ch==0:
list1.append(ch)
if len(list1)==2:
count += 1
print(i,end=", ")
print()
return count
num2 = int(input("enter a number: "))
result=findprime(num2)
print("prime numbers between 1 and",num2,"are",result)
回答11:
Here's a more extensive example with optimization in mind for Python 3.
import sys
inner_loop_iterations: int = 0
def is_prime(n):
a: int = 2
global inner_loop_iterations
if n == 1:
return("Not prime")
elif n == 2:
return("Prime")
while a * a <= n + 1:
inner_loop_iterations += 1
# This if statement reduces the number of inner loop iterations by roughy 50%
# just weeding out the even numbers.
if a % 2 == 0:
a += 1
else:
a += 2
if n % 2 == 0 or n % a == 0:
return ("Not prime")
else:
return ("Prime")
while True:
sys.stdout.write("Enter number to see if it's prime ('q' to quit): ")
n = input()
if not n:
continue
if n == 'q':
break
try:
n = int(n)
except ValueError:
print("Please enter a valid number")
if n < 1:
print("Please enter a valid number")
continue
sys.stdout.write("{}\n".format(is_prime(n)))
sys.stderr.write("Inner loops: {}\n\n".format(inner_loop_iterations))
inner_loop_iterations=0
This program has two main optimizations, first it only iterates from 2 to the square root of n and it only iterates through odd numbers. Using these optimizations I was able to find out that the number 1000000007 is prime in only 15811 loop iterations.
回答12:
Better use
for i in range(2, p//2 + 1):
回答13:
This in my opinion is a more optimised way. This finds all the prime numbers up to 1,000,000 in less than 8 seconds on my setup.
It is also one of my very first attempts at python, so I stand to be corrected
class prime:
def finder (self):
import math
n = long(raw_input("What number should I go up to? "))
for i in range(2, n):
is_prime = True
if i % 2 == 0:
is_prime = False
for j in range(3, long(math.sqrt(i) + 1), 2):
if i % j == 0:
is_prime = False
break
if is_prime:
print(i)
prime().finder()
来源：https://stackoverflow.com/questions/14656473/python-beginners-loop-finding-primes