+ 2
How do I iterate between 5 digits and return the 4 digits without duplicated digits
Input_unit = [1,2,3,4,5] Output_digit = 1,2,3,4 1,2,3,5 1,2,4,3 1,2,4,5 1,2,5,3 1,2,5,4 . . . No two digits must be unique in output result
4 ответов
+ 3
Things like this can be done with module itertools and function permutations.
permutations(<sequence>, count)
<sequence> can be a list with all input characters (in your case: 1,2,3,4,5), count defines the number of characters used for each output the permutations generates. With your settings it eill generate 120 different outputs.
The output is a permutation object, that can be converted to a list.
+ 2
Yes you can, but I am afraid these versions will be more difficult to understand than itertools with permutation. You can use:
- nested loops
- recursion
+ 1
Thanks Lothar... I'm a phyton beginner. Can i achieve this without using module?
+ 1
Here it is
b=[1,2,3,4,5]
a=list((i,j,k,l) for i in b for j in b for k in b for l in b)
def appearance(i):
b={}
count=0
for i in a:
if i not in b:
count+=1
b[i]=count
elif i in b:
for l,m in b.items():
m+=1
b[l]=m
m=0
count=0
if max(b.values())>1:
return 2
else:
return 1
for i in a:
if appearance(i)==1:
for j in str(i):
if j!="(" and j!=")":
print(j,end="")
print("")