Replace min & max value in a list

-3

I need help to write a python function for lists that will do two things.

First instance, it will replace the largest and smallest elements in the list with the number 5000 (for largest) and -5000 (for smallest).

thelist = input("Please input a list with numbers: ")

mylist = list(map(int, thelist.split()))

Please input a list with numbers: 1 2 3 4 5 6 7 8 9 10

print(mylist)

Result: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Honestly I have no clue how to replace the 1 and 10 with the numbers 5000 and -5000 using a py function. I can get the min and max number but replacing them is something I don't know how to do.

edited Nov 20 at 8:16

Elisha

19.5k45069

asked Nov 20 at 7:15

blargh

1

mylist[0] and mylist[10] will let you access the first and last members of the list. len(mylist) will give you the length of the list.
– Soumya Kanti
Nov 20 at 7:20

add a comment |

-3

I need help to write a python function for lists that will do two things.

First instance, it will replace the largest and smallest elements in the list with the number 5000 (for largest) and -5000 (for smallest).

thelist = input("Please input a list with numbers: ")

mylist = list(map(int, thelist.split()))

Please input a list with numbers: 1 2 3 4 5 6 7 8 9 10

print(mylist)

Result: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Honestly I have no clue how to replace the 1 and 10 with the numbers 5000 and -5000 using a py function. I can get the min and max number but replacing them is something I don't know how to do.

edited Nov 20 at 8:16

Elisha

19.5k45069

asked Nov 20 at 7:15

blargh

1

mylist[0] and mylist[10] will let you access the first and last members of the list. len(mylist) will give you the length of the list.
– Soumya Kanti
Nov 20 at 7:20

add a comment |

-3

I need help to write a python function for lists that will do two things.

First instance, it will replace the largest and smallest elements in the list with the number 5000 (for largest) and -5000 (for smallest).

thelist = input("Please input a list with numbers: ")

mylist = list(map(int, thelist.split()))

Please input a list with numbers: 1 2 3 4 5 6 7 8 9 10

print(mylist)

Result: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Honestly I have no clue how to replace the 1 and 10 with the numbers 5000 and -5000 using a py function. I can get the min and max number but replacing them is something I don't know how to do.

edited Nov 20 at 8:16

Elisha

19.5k45069

asked Nov 20 at 7:15

blargh

I need help to write a python function for lists that will do two things.

First instance, it will replace the largest and smallest elements in the list with the number 5000 (for largest) and -5000 (for smallest).

thelist = input("Please input a list with numbers: ")

mylist = list(map(int, thelist.split()))

Please input a list with numbers: 1 2 3 4 5 6 7 8 9 10

print(mylist)

Result: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Honestly I have no clue how to replace the 1 and 10 with the numbers 5000 and -5000 using a py function. I can get the min and max number but replacing them is something I don't know how to do.

python list

edited Nov 20 at 8:16

Elisha

19.5k45069

asked Nov 20 at 7:15

blargh

edited Nov 20 at 8:16

Elisha

19.5k45069

asked Nov 20 at 7:15

blargh

edited Nov 20 at 8:16

Elisha

19.5k45069

edited Nov 20 at 8:16

Elisha

19.5k45069

edited Nov 20 at 8:16

Elisha

19.5k45069

asked Nov 20 at 7:15

blargh

asked Nov 20 at 7:15

blargh

asked Nov 20 at 7:15

blargh

1

mylist[0] and mylist[10] will let you access the first and last members of the list. len(mylist) will give you the length of the list.
– Soumya Kanti
Nov 20 at 7:20

add a comment |

1

mylist[0] and mylist[10] will let you access the first and last members of the list. len(mylist) will give you the length of the list.
– Soumya Kanti
Nov 20 at 7:20

mylist[0] and mylist[10] will let you access the first and last members of the list. len(mylist) will give you the length of the list.
– Soumya Kanti
Nov 20 at 7:20

add a comment |

5 Answers
5

active

oldest

votes

In a bit condensed format using list comprehension, You can use

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

b = [-5000 if x == min(a) else 5000 if x == max(a) else x for x in a]

It replaces the minimal elements by -5000 and the maximal ones by 5000. f there are several minimal values, all of them will be replaced. You can use a function to use other values than -5000, 5000.

answered Nov 20 at 7:24

Sam The Sid

216

add a comment |

You can use mylist[0] = -5000 and mylist[-1] = 5000.

The negative index counts from the right side of the list

Or if your list is jumbled

Try this,

min_pos = mylist.index(min(mylist))

max_pos = mylist.index(max(mylist))



mylist[min_pos] = -5000

mylist[max_pos] = 5000



print(mylist)

Hope this answers your question

answered Nov 20 at 7:31

Avichal Bettoli

493

add a comment |

mylist[mylist.index(min(mylist))] = -5000

mylist[mylist.index(max(mylist))] = 5000

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

Thank you very much Rudolf! I created a function that utilized this and it replaced the min/max with the -5000 and 5000. Thank you very much! I will further study this index feature, had no clue it was even a thing!
– blargh
Nov 20 at 7:30

Glad I could help you)
– Rudolf Morkovskyi
Nov 20 at 7:34

add a comment |

This can be done by two phases over min/max:

arr = [1,2,3,4]

arr[arr.index(max(arr))] = 5000

This yields:

[1, 2, 3, 5000]

The two phases here are:

Find the max value: max(arr). In this example, the value will be 4.

Find the index of max value: arr.index(4). In this example, the index will be 3.

The two phases lead to assignment at the index at the max value cell. Note that this assumes a unique max value.

edited Nov 20 at 7:30

answered Nov 20 at 7:21

Elisha

19.5k45069

add a comment |

You can have a function like this:

In [361]: def replace_vals(l):

     ...:     min_index = l.index(min(l))  # min(l) finds minimum of list. 'l.index()' finds the index of a given value.

     ...:     max_index = l.index(max(l))  # max(l) finds maximum of list.

     ...:     l[min_index] = -5000 # Just replace min_index found above by -5000

     ...:     l[max_index] = 5000  # Just replace max_index found above by 5000

     ...:     

In [355]: lst=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]



In [362]: replace_vals(lst)



In [363]: lst

Out[363]: [-5000, 2, 3, 4, 5, 6, 7, 8, 9, 5000]

edited Nov 20 at 7:31

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53387996%2freplace-min-max-value-in-a-list%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

In a bit condensed format using list comprehension, You can use

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

b = [-5000 if x == min(a) else 5000 if x == max(a) else x for x in a]

It replaces the minimal elements by -5000 and the maximal ones by 5000. f there are several minimal values, all of them will be replaced. You can use a function to use other values than -5000, 5000.

answered Nov 20 at 7:24

Sam The Sid

216

add a comment |

In a bit condensed format using list comprehension, You can use

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

b = [-5000 if x == min(a) else 5000 if x == max(a) else x for x in a]

It replaces the minimal elements by -5000 and the maximal ones by 5000. f there are several minimal values, all of them will be replaced. You can use a function to use other values than -5000, 5000.

answered Nov 20 at 7:24

Sam The Sid

216

add a comment |

In a bit condensed format using list comprehension, You can use

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

b = [-5000 if x == min(a) else 5000 if x == max(a) else x for x in a]

It replaces the minimal elements by -5000 and the maximal ones by 5000. f there are several minimal values, all of them will be replaced. You can use a function to use other values than -5000, 5000.

answered Nov 20 at 7:24

Sam The Sid

216

In a bit condensed format using list comprehension, You can use

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

b = [-5000 if x == min(a) else 5000 if x == max(a) else x for x in a]

It replaces the minimal elements by -5000 and the maximal ones by 5000. f there are several minimal values, all of them will be replaced. You can use a function to use other values than -5000, 5000.

answered Nov 20 at 7:24

Sam The Sid

216

answered Nov 20 at 7:24

Sam The Sid

216

answered Nov 20 at 7:24

Sam The Sid

216

answered Nov 20 at 7:24

Sam The Sid

216

add a comment |

You can use mylist[0] = -5000 and mylist[-1] = 5000.

The negative index counts from the right side of the list

Or if your list is jumbled

Try this,

min_pos = mylist.index(min(mylist))

max_pos = mylist.index(max(mylist))



mylist[min_pos] = -5000

mylist[max_pos] = 5000



print(mylist)

Hope this answers your question

answered Nov 20 at 7:31

Avichal Bettoli

493

add a comment |

You can use mylist[0] = -5000 and mylist[-1] = 5000.

The negative index counts from the right side of the list

Or if your list is jumbled

Try this,

min_pos = mylist.index(min(mylist))

max_pos = mylist.index(max(mylist))



mylist[min_pos] = -5000

mylist[max_pos] = 5000



print(mylist)

Hope this answers your question

answered Nov 20 at 7:31

Avichal Bettoli

493

add a comment |

You can use mylist[0] = -5000 and mylist[-1] = 5000.

The negative index counts from the right side of the list

Or if your list is jumbled

Try this,

min_pos = mylist.index(min(mylist))

max_pos = mylist.index(max(mylist))



mylist[min_pos] = -5000

mylist[max_pos] = 5000



print(mylist)

Hope this answers your question

answered Nov 20 at 7:31

Avichal Bettoli

493

You can use mylist[0] = -5000 and mylist[-1] = 5000.

The negative index counts from the right side of the list

Or if your list is jumbled

Try this,

min_pos = mylist.index(min(mylist))

max_pos = mylist.index(max(mylist))



mylist[min_pos] = -5000

mylist[max_pos] = 5000



print(mylist)

Hope this answers your question

answered Nov 20 at 7:31

Avichal Bettoli

493

answered Nov 20 at 7:31

Avichal Bettoli

493

answered Nov 20 at 7:31

Avichal Bettoli

493

answered Nov 20 at 7:31

Avichal Bettoli

493

add a comment |

mylist[mylist.index(min(mylist))] = -5000

mylist[mylist.index(max(mylist))] = 5000

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

Thank you very much Rudolf! I created a function that utilized this and it replaced the min/max with the -5000 and 5000. Thank you very much! I will further study this index feature, had no clue it was even a thing!
– blargh
Nov 20 at 7:30

Glad I could help you)
– Rudolf Morkovskyi
Nov 20 at 7:34

add a comment |

mylist[mylist.index(min(mylist))] = -5000

mylist[mylist.index(max(mylist))] = 5000

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

Thank you very much Rudolf! I created a function that utilized this and it replaced the min/max with the -5000 and 5000. Thank you very much! I will further study this index feature, had no clue it was even a thing!
– blargh
Nov 20 at 7:30

Glad I could help you)
– Rudolf Morkovskyi
Nov 20 at 7:34

add a comment |

mylist[mylist.index(min(mylist))] = -5000

mylist[mylist.index(max(mylist))] = 5000

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

mylist[mylist.index(min(mylist))] = -5000

mylist[mylist.index(max(mylist))] = 5000

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

answered Nov 20 at 7:22

Rudolf Morkovskyi

720117

Thank you very much Rudolf! I created a function that utilized this and it replaced the min/max with the -5000 and 5000. Thank you very much! I will further study this index feature, had no clue it was even a thing!
– blargh
Nov 20 at 7:30

Glad I could help you)
– Rudolf Morkovskyi
Nov 20 at 7:34

add a comment |

Thank you very much Rudolf! I created a function that utilized this and it replaced the min/max with the -5000 and 5000. Thank you very much! I will further study this index feature, had no clue it was even a thing!
– blargh
Nov 20 at 7:30

Glad I could help you)
– Rudolf Morkovskyi
Nov 20 at 7:34

Thank you very much Rudolf! I created a function that utilized this and it replaced the min/max with the -5000 and 5000. Thank you very much! I will further study this index feature, had no clue it was even a thing!
– blargh
Nov 20 at 7:30

Glad I could help you)
– Rudolf Morkovskyi
Nov 20 at 7:34

add a comment |

This can be done by two phases over min/max:

arr = [1,2,3,4]

arr[arr.index(max(arr))] = 5000

This yields:

[1, 2, 3, 5000]

The two phases here are:

Find the max value: max(arr). In this example, the value will be 4.

Find the index of max value: arr.index(4). In this example, the index will be 3.

The two phases lead to assignment at the index at the max value cell. Note that this assumes a unique max value.

edited Nov 20 at 7:30

answered Nov 20 at 7:21

Elisha

19.5k45069

add a comment |

This can be done by two phases over min/max:

arr = [1,2,3,4]

arr[arr.index(max(arr))] = 5000

This yields:

[1, 2, 3, 5000]

The two phases here are:

Find the max value: max(arr). In this example, the value will be 4.

Find the index of max value: arr.index(4). In this example, the index will be 3.

The two phases lead to assignment at the index at the max value cell. Note that this assumes a unique max value.

edited Nov 20 at 7:30

answered Nov 20 at 7:21

Elisha

19.5k45069

add a comment |

This can be done by two phases over min/max:

arr = [1,2,3,4]

arr[arr.index(max(arr))] = 5000

This yields:

[1, 2, 3, 5000]

The two phases here are:

Find the max value: max(arr). In this example, the value will be 4.

Find the index of max value: arr.index(4). In this example, the index will be 3.

The two phases lead to assignment at the index at the max value cell. Note that this assumes a unique max value.

edited Nov 20 at 7:30

answered Nov 20 at 7:21

Elisha

19.5k45069

This can be done by two phases over min/max:

arr = [1,2,3,4]

arr[arr.index(max(arr))] = 5000

This yields:

[1, 2, 3, 5000]

The two phases here are:

Find the max value: max(arr). In this example, the value will be 4.

Find the index of max value: arr.index(4). In this example, the index will be 3.

The two phases lead to assignment at the index at the max value cell. Note that this assumes a unique max value.

edited Nov 20 at 7:30

answered Nov 20 at 7:21

Elisha

19.5k45069

edited Nov 20 at 7:30

answered Nov 20 at 7:21

Elisha

19.5k45069

answered Nov 20 at 7:21

Elisha

19.5k45069

answered Nov 20 at 7:21

Elisha

19.5k45069

add a comment |

You can have a function like this:

In [361]: def replace_vals(l):

     ...:     min_index = l.index(min(l))  # min(l) finds minimum of list. 'l.index()' finds the index of a given value.

     ...:     max_index = l.index(max(l))  # max(l) finds maximum of list.

     ...:     l[min_index] = -5000 # Just replace min_index found above by -5000

     ...:     l[max_index] = 5000  # Just replace max_index found above by 5000

     ...:     

In [355]: lst=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]



In [362]: replace_vals(lst)



In [363]: lst

Out[363]: [-5000, 2, 3, 4, 5, 6, 7, 8, 9, 5000]

edited Nov 20 at 7:31

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

add a comment |

You can have a function like this:

In [361]: def replace_vals(l):

     ...:     min_index = l.index(min(l))  # min(l) finds minimum of list. 'l.index()' finds the index of a given value.

     ...:     max_index = l.index(max(l))  # max(l) finds maximum of list.

     ...:     l[min_index] = -5000 # Just replace min_index found above by -5000

     ...:     l[max_index] = 5000  # Just replace max_index found above by 5000

     ...:     

In [355]: lst=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]



In [362]: replace_vals(lst)



In [363]: lst

Out[363]: [-5000, 2, 3, 4, 5, 6, 7, 8, 9, 5000]

edited Nov 20 at 7:31

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

add a comment |

You can have a function like this:

In [361]: def replace_vals(l):

     ...:     min_index = l.index(min(l))  # min(l) finds minimum of list. 'l.index()' finds the index of a given value.

     ...:     max_index = l.index(max(l))  # max(l) finds maximum of list.

     ...:     l[min_index] = -5000 # Just replace min_index found above by -5000

     ...:     l[max_index] = 5000  # Just replace max_index found above by 5000

     ...:     

In [355]: lst=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]



In [362]: replace_vals(lst)



In [363]: lst

Out[363]: [-5000, 2, 3, 4, 5, 6, 7, 8, 9, 5000]

edited Nov 20 at 7:31

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

You can have a function like this:

In [361]: def replace_vals(l):

     ...:     min_index = l.index(min(l))  # min(l) finds minimum of list. 'l.index()' finds the index of a given value.

     ...:     max_index = l.index(max(l))  # max(l) finds maximum of list.

     ...:     l[min_index] = -5000 # Just replace min_index found above by -5000

     ...:     l[max_index] = 5000  # Just replace max_index found above by 5000

     ...:     

In [355]: lst=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]



In [362]: replace_vals(lst)



In [363]: lst

Out[363]: [-5000, 2, 3, 4, 5, 6, 7, 8, 9, 5000]

edited Nov 20 at 7:31

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

edited Nov 20 at 7:31

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

answered Nov 20 at 7:22

Mayank Porwal

4,4541623

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Argthtjtr