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.






python list







share|improve this question




















  • 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
















-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.






python list







share|improve this question




















  • 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














-3












-3








-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.






python list







share|improve this question















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




python list






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 at 8:16









Elisha

19.5k45069




19.5k45069










asked Nov 20 at 7:15









blargh

13




13








  • 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














  • 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








1




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












5 Answers
5






active

oldest

votes


















1














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.






share|improve this answer





























    1














    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






    share|improve this answer





























      0














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





      share|improve this answer





















      • 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



















      0














      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:




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

      2. 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.






      share|improve this answer































        0














        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]





        share|improve this answer























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          5 Answers
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          5 Answers
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          active

          oldest

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          1














          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.






          share|improve this answer


























            1














            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.






            share|improve this answer
























              1












              1








              1






              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.






              share|improve this answer












              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.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 20 at 7:24









              Sam The Sid

              216




              216

























                  1














                  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






                  share|improve this answer


























                    1














                    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






                    share|improve this answer
























                      1












                      1








                      1






                      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






                      share|improve this answer












                      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







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Nov 20 at 7:31









                      Avichal Bettoli

                      493




                      493























                          0














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





                          share|improve this answer





















                          • 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
















                          0














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





                          share|improve this answer





















                          • 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














                          0












                          0








                          0






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





                          share|improve this answer












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






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 20 at 7:22









                          Rudolf Morkovskyi

                          720117




                          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


















                          • 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




                          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




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











                          0














                          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:




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

                          2. 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.






                          share|improve this answer




























                            0














                            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:




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

                            2. 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.






                            share|improve this answer


























                              0












                              0








                              0






                              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:




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

                              2. 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.






                              share|improve this answer














                              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:




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

                              2. 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.







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited Nov 20 at 7:30

























                              answered Nov 20 at 7:21









                              Elisha

                              19.5k45069




                              19.5k45069























                                  0














                                  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]





                                  share|improve this answer




























                                    0














                                    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]





                                    share|improve this answer


























                                      0












                                      0








                                      0






                                      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]





                                      share|improve this answer














                                      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]






                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited Nov 20 at 7:31

























                                      answered Nov 20 at 7:22









                                      Mayank Porwal

                                      4,4541623




                                      4,4541623






























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                                          Can an atomic nucleus contain both particles and antiparticles? [duplicate]

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                                          9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
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