如何检查列表中的数字而不检测具有相同数字的数字？

Question

Say I have a list: [2 dogs play, 4 dogs play, 22 dogs play, 24 dogs play, 26 dogs play] I have an expression that asks a user for a number and it's stored in the variable, num

I have a condition in my code where,

for item in list:
     if num in item:
        ....
        do something to item

My problem is that if the user inputs 2, the code will also do something to the items that have 22, 24, and 26 because it has 2 in it, when I only want it to do something to the item that has only 2. How do I accomplish this?

Answer 1

Solution:

Change your function to:

for item in list:
    #  Since it's obvious you're checking a string for a number (in string form), you need 
    #  to make sure a space comes after the number (at the beginning of the string) in order
    #  to avoid incomplete matches.
    if item.startswith(num, beg=0, end=len(num)):
        ...
        func_do(item)

Results:

A)

['2 dogs play', '4 dogs play', '22 dogs play', '24 dogs play', '26 dogs play']
num = '2' #  input with no space trailing

Output using this method is the result of func_do('2 dogs play')

B)

['2 dogs play', '4 dogs play', '22 dogs play', '24 dogs play', '26 dogs play']
num = '2 ' #  input with space trailing (user hit spacebar and we didn't `trim()` the input)

Output using this method is still the result of func_do('2 dogs play')

Beware:

Sanitize your input, if you use any other method provided so far (or any method that checks for spaces following the input) you will have to be wary of a user entering a number with a space after it (or before it).

Use num = input().strip() or:

num = input()
num = num.strip()

ALSO: This answer also obviously relies on the user-input string that you're trying to match residing at the beginning of the item strings from your list .

Answer 2

You need to get the digits from the item and then check if it is equal to num :

Using regex:

import re

uList = ['2 dogs play', '4 dogs play', '22 dogs play', '24 dogs play', '26 dogs play']
num = int(input("Enter a num: "))

for item in uList:
      if num == int((re.findall(r'\d+', item))[0]):
         print(item)

OUTPUT :

Enter a num: 2
2 dogs play

Process finished with exit code 0

EDIT :

Using split() :

uList = ['2 dogs play', '4 dogs play', '22 dogs play', '24 dogs play', '26 dogs play']
num = input("Enter a num: ")   # no conversion to int here

for item in uList:
     if num == item.split(' ')[0]:  # split and get the digits before space
         print(item)

OUTPUT :

Enter a num: 22
22 dogs play

Process finished with exit code 0

Answer 3

There will be many ways to solve this problem, one way using regex is :

import re

for item in a:
   if re.search("^"+num+" ", item):
        print(item) 
        # Cool logic goes here 

Answer 4

 num= raw_input("enter the number that you want to check")
   if(num==item.split(' ', 1)[0])
   ...do something

that will check if the number exists in the list given that the number comes first in the list
If you want to explicitly compare it to an int then the other provided answers are what you are looking for.

Answer 5

I took the super simple approach of splitting each item in the list and querying the first.

num = 2

list = ['2 dogs play', '4 dogs play', '22 dogs play', '24 dogs play', '26 dogs play']

for item in list:
    number_of_dogs = int(item.split(' ')[0])
    if number_of_dogs == num:
        print('Doing something!')

The code is very simplistic and only works if each item in the list starts with the number followed by a space.