Python数学代码大于和小于不起作用

Question

import numpy as np
import random
i = random.randint(1,100)
if i > 0 and i <16:
    print ("Broken")
else if i > 16 and i > 100
    print ("Not broken")

I am trying to make it if the number is between 1 to 15 the racquet is broken but if it is 16-100 it is not broken. It says it is invalid syntax in python. Why is it invalid syntax?

Answer 1

You've got 2 SyntaxErrors:

• There is no else if in Python, just elif

• You need a : after the conditions .

And a logical mistake:

• i > 100 can never be True .

Then you also don't need and here, you could just use:

import random
i = random.randint(1,100)
if i < 16:
    print ("Broken")
else:
    print ("Not broken")

There is also the shortened version of i > 0 and i < 16 :

if 0 < i < 16:
    print ("Broken")
elif 16 < i <= 100:  # compare to "<= 100" instead of "> 100"
    print ("Not broken")

Answer 2

You appear to be trying to "fold" a nested if statement into its containing if statement, C-style:

if (i > 0 && i < 16)
    printf("Broken\n");
else
    if (i > 16 && i < 100)
        printf("Not broken\n");

Because the whitespace isn't significant, the above is equivalent to

if (i > 0 && i < 16)
    printf("Broken\n");
else if (i > 16 && i < 100)
    printf("Not broken\n");

giving the illusion of an else if clause.

---

In Python, indentation is significant, so you can't pull the same trick as in C. Instead, Python has an explicit elif clause that you can use to check multiple conditions in a single if statement.

if i > 0 and i < 16:
    print("Broken")
elif i > 16 and i < 100:
    print("Not broken")

This is semantically equivalent to

if i > 0 and i < 16:
    print("Broken")
else:
    if i > 16 and i < 100:
        print("Not broken")

but nicer looking.

Answer 3

In Python, you use elif instead of else if . You also need a colon at the end of your else if line.

Answer 4

It should be:

elif i > 16 and i > 100:

elif and with ":" in the end