Check if an input is equal to any item in the list [duplicate]

This question already has an answer here:

Fastest way to check if a value exist in a list

12 answers

let's say we have a list

a=["a1", "a2", "a3"]

then we have an input:

x=input("enter something: ")

how would you check if that input "x" is equal to any item in the list and then return true?

asked Nov 22 at 18:05

Coder Cody

408

marked as duplicate by slider, quamrana, Austin, Rahul K P, Mr. T Nov 22 at 20:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can do if x in a: return True.
– slider
Nov 22 at 18:06

it is not a duplicate of that article as i have a different situation
– Coder Cody
Nov 22 at 18:12

How is it different? Can you be more specific?
– slider
Nov 22 at 18:14

I didn't really want to disclose this information but I'm trying to make a naughts and crosses game and I've created lists to hold the data of each square in the naughts and crosses. I will then ask for an input and use that input, not to count how many times it repeats but use it to simply see whether the input correlates to a keyword to describe each square. I just did not find my answer on that particular post
– Coder Cody
Nov 22 at 18:17

1

@slider Just x in a will do the trick.
– Klaus D.
Nov 22 at 19:11

add a comment |

This question already has an answer here:

Fastest way to check if a value exist in a list

12 answers

let's say we have a list

a=["a1", "a2", "a3"]

then we have an input:

x=input("enter something: ")

how would you check if that input "x" is equal to any item in the list and then return true?

asked Nov 22 at 18:05

Coder Cody

408

marked as duplicate by slider, quamrana, Austin, Rahul K P, Mr. T Nov 22 at 20:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can do if x in a: return True.
– slider
Nov 22 at 18:06

it is not a duplicate of that article as i have a different situation
– Coder Cody
Nov 22 at 18:12

How is it different? Can you be more specific?
– slider
Nov 22 at 18:14

I didn't really want to disclose this information but I'm trying to make a naughts and crosses game and I've created lists to hold the data of each square in the naughts and crosses. I will then ask for an input and use that input, not to count how many times it repeats but use it to simply see whether the input correlates to a keyword to describe each square. I just did not find my answer on that particular post
– Coder Cody
Nov 22 at 18:17

1

@slider Just x in a will do the trick.
– Klaus D.
Nov 22 at 19:11

add a comment |

This question already has an answer here:

Fastest way to check if a value exist in a list

12 answers

let's say we have a list

a=["a1", "a2", "a3"]

then we have an input:

x=input("enter something: ")

how would you check if that input "x" is equal to any item in the list and then return true?

asked Nov 22 at 18:05

Coder Cody

408

This question already has an answer here:

Fastest way to check if a value exist in a list

12 answers

let's say we have a list

a=["a1", "a2", "a3"]

then we have an input:

x=input("enter something: ")

how would you check if that input "x" is equal to any item in the list and then return true?

This question already has an answer here:

Fastest way to check if a value exist in a list

12 answers

python

asked Nov 22 at 18:05

Coder Cody

408

asked Nov 22 at 18:05

Coder Cody

408

asked Nov 22 at 18:05

Coder Cody

408

asked Nov 22 at 18:05

Coder Cody

408

asked Nov 22 at 18:05

Coder Cody

408

marked as duplicate by slider, quamrana, Austin, Rahul K P, Mr. T Nov 22 at 20:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by slider, quamrana, Austin, Rahul K P, Mr. T Nov 22 at 20:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can do if x in a: return True.
– slider
Nov 22 at 18:06

it is not a duplicate of that article as i have a different situation
– Coder Cody
Nov 22 at 18:12

How is it different? Can you be more specific?
– slider
Nov 22 at 18:14

I didn't really want to disclose this information but I'm trying to make a naughts and crosses game and I've created lists to hold the data of each square in the naughts and crosses. I will then ask for an input and use that input, not to count how many times it repeats but use it to simply see whether the input correlates to a keyword to describe each square. I just did not find my answer on that particular post
– Coder Cody
Nov 22 at 18:17

1

@slider Just x in a will do the trick.
– Klaus D.
Nov 22 at 19:11

add a comment |

You can do if x in a: return True.
– slider
Nov 22 at 18:06

it is not a duplicate of that article as i have a different situation
– Coder Cody
Nov 22 at 18:12

How is it different? Can you be more specific?
– slider
Nov 22 at 18:14

I didn't really want to disclose this information but I'm trying to make a naughts and crosses game and I've created lists to hold the data of each square in the naughts and crosses. I will then ask for an input and use that input, not to count how many times it repeats but use it to simply see whether the input correlates to a keyword to describe each square. I just did not find my answer on that particular post
– Coder Cody
Nov 22 at 18:17

1

@slider Just x in a will do the trick.
– Klaus D.
Nov 22 at 19:11

You can do if x in a: return True.
– slider
Nov 22 at 18:06

it is not a duplicate of that article as i have a different situation
– Coder Cody
Nov 22 at 18:12

How is it different? Can you be more specific?
– slider
Nov 22 at 18:14

I didn't really want to disclose this information but I'm trying to make a naughts and crosses game and I've created lists to hold the data of each square in the naughts and crosses. I will then ask for an input and use that input, not to count how many times it repeats but use it to simply see whether the input correlates to a keyword to describe each square. I just did not find my answer on that particular post
– Coder Cody
Nov 22 at 18:17

@slider Just x in a will do the trick.
– Klaus D.
Nov 22 at 19:11

add a comment |

1 Answer
1

active

oldest

votes

You can check it with that code:

print("Yes, x in list" if x in a else "No, x is not in list")

It same that:

if x in a:

    print("Yes, x in list")

else:

    print("No, x is not in list")

Also if you want to get index of contained element - just use .index():

if x in a: 

    print('Yes, x in list, and it index is:', a.index(x))

In that example i'am avoid else: construct, but you can use it like in previous example, if you need it.

edited Nov 22 at 18:24

answered Nov 22 at 18:12

Andrey Suglobov

1248

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

You can check it with that code:

print("Yes, x in list" if x in a else "No, x is not in list")

It same that:

if x in a:

    print("Yes, x in list")

else:

    print("No, x is not in list")

Also if you want to get index of contained element - just use .index():

if x in a: 

    print('Yes, x in list, and it index is:', a.index(x))

In that example i'am avoid else: construct, but you can use it like in previous example, if you need it.

edited Nov 22 at 18:24

answered Nov 22 at 18:12

Andrey Suglobov

1248

add a comment |

You can check it with that code:

print("Yes, x in list" if x in a else "No, x is not in list")

It same that:

if x in a:

    print("Yes, x in list")

else:

    print("No, x is not in list")

Also if you want to get index of contained element - just use .index():

if x in a: 

    print('Yes, x in list, and it index is:', a.index(x))

In that example i'am avoid else: construct, but you can use it like in previous example, if you need it.

edited Nov 22 at 18:24

answered Nov 22 at 18:12

Andrey Suglobov

1248

add a comment |

You can check it with that code:

print("Yes, x in list" if x in a else "No, x is not in list")

It same that:

if x in a:

    print("Yes, x in list")

else:

    print("No, x is not in list")

Also if you want to get index of contained element - just use .index():

if x in a: 

    print('Yes, x in list, and it index is:', a.index(x))

In that example i'am avoid else: construct, but you can use it like in previous example, if you need it.

edited Nov 22 at 18:24

answered Nov 22 at 18:12

Andrey Suglobov

1248

You can check it with that code:

print("Yes, x in list" if x in a else "No, x is not in list")

It same that:

if x in a:

    print("Yes, x in list")

else:

    print("No, x is not in list")

Also if you want to get index of contained element - just use .index():

if x in a: 

    print('Yes, x in list, and it index is:', a.index(x))

In that example i'am avoid else: construct, but you can use it like in previous example, if you need it.

edited Nov 22 at 18:24

answered Nov 22 at 18:12

Andrey Suglobov

1248

edited Nov 22 at 18:24

answered Nov 22 at 18:12

Andrey Suglobov

1248

answered Nov 22 at 18:12

Andrey Suglobov

1248

answered Nov 22 at 18:12

Andrey Suglobov

1248

add a comment |

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