Python-检查一个数字，看看它是否是使用除数的素数

Question

我必须编写一个程序，通过遍历范围（1，num）中的除数来检查数字是否为素数。 （例如，如果输入为5，程序将检查2是否为除数，然后是3，然后是4）。有人可以解释我如何将最后一个表达式放在我的程序中（“这是一个素数”打印函数），这样它只出现在“for”循环中的“else”函数之后而不是“if div = = 0：“陈述？谢谢！

while True:
    num = int(input("Enter a positive number to see if it's a prime number: "))
    if num > 1:
        break
    elif num == 1:
        print("1 is technically not a prime number.")
    else:
        print("Number cannot be negative- try again.")

for num1 in range(2,num):
    div = num%num1
    if div == 0:
        print(num1,"is a divisor of",num,"... stopping.")
        print(num,"is not a prime number.")
        break
    else:
        print(num1,"is NOT a divisor of",num,"... continuing")

print(num,"is a prime number!")

Answer 1

有很多方法可以做到这一点：

设置标志！

is_prime = True # a boolean flag
for num1 in range(2,num):
    div = num%num1
    if div == 0:
        print(num1,"is a divisor of",num,"... stopping.")
        print(num,"is not a prime number.")
        is_prime = False
        break
    else:
        print(num1,"is NOT a divisor of",num,"... continuing")

 if is_prime:
     print(num,"is a prime number!")

使用for-else！

for num1 in range(2,num):
    div = num%num1
    if div == 0:
        print(num1,"is a divisor of",num,"... stopping.")
        print(num,"is not a prime number.")
        is_prime = False
        break
    else:
        print(num1,"is NOT a divisor of",num,"... continuing")
 else: 
      print(num,"is a prime number!") # will execute only when for-loop ends

Answer 2

Python的for表达式有an optional else clause：

  

循环语句可能有一个else子句;当循环通过列表耗尽（with for）或条件变为false（with while）时终止，但是当循环被break语句终止时，执行它。

所以你的代码应该是这样的：

while True:
    num = int(input("Enter a positive number to see if it's a prime number: "))
    if num > 1:
        break
    elif num == 1:
        print("1 is technically not a prime number.")
    else:
        print("Number cannot be negative- try again.")

for num1 in range(2,num):
    div = num%num1
    if div == 0:
        print(num1,"is a divisor of",num,"... stopping.")
        print(num,"is not a prime number.")
        break
    else:
        print(num1,"is NOT a divisor of",num,"... continuing")
else:
    print(num,"is a prime number!")

Answer 3

鉴于你已经将除法除以1，所有你需要做的就是计算是否只有一个除数（数字本身）。因此，为了保持代码结构，我添加了一个计数器。

while True:
    num = int(input("Enter a positive number to see if it's a prime number: "))
    if num > 1:
        break
    elif num == 1:
        print("1 is technically not a prime number.")
    else:
        print("Number cannot be negative- try again.")

for num1 in range(2,num):
    div = num%num1
    count = 0
    if div == 0:
        print(num1,"is a divisor of",num,"... stopping.")
        print(num,"is not a prime number.")
        break
    else:
        count += 1
        print(num1,"is NOT a divisor of",num,"... continuing")

if count == 1:
    print(num,"is a prime number!")

Answer 4

我添加了一行sys.exit（）

library(dplyr)

df <- read.table(text = "RT   Max    blk24  blk23    mat23   sm23    sm24
    1   1.6  261    13     19       19      27      20
    2   1.6  284    NA     53       5       99      91
    3   1.7  304    NA     NA       NA      81      NA
    4   1.8  305    32     28       28      39      33
    5   1.9  322    8      NA       NA      36      39
    6   1.9  349    NA     NA       NA      24      17
    7   2    389    1       3       NA      50      NA
    8   2.1  393    80      70      7       12      71
    9   2.1  622    NA      NA      15      97      96
    10  2.2  705    NA      NA      NA      NA      32", header = T)

nams <- df %>% select(matches('blk|mat')) 
df <- df[apply(nams, 1, function(i) all(is.na(i)))==TRUE,]

df

    RT Max blk24 blk23 mat23 sm23 sm24
3  1.7 304    NA    NA    NA   81   NA
6  1.9 349    NA    NA    NA   24   17
10 2.2 705    NA    NA    NA   NA   32