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Commit 03b41ba

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‎Itertools/Compress the String!.py‎

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from itertools import groupby
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ls=list(map(int, " ".join(input()).split(" ")))
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for key, group in(groupby(ls)):
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print((len((list(group))), key),end=" ")
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from itertools import combinations
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st,*r = input().split()
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for i in range(1,int(r[0])+1):
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for j in list(combinations(sorted(st),i)):
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print(*j,sep="")
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# these are the more ways to find permutations
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#>>> from itertools import combinations
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#>>>
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#>>> print list(combinations('12345',2))
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#[('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '4'), ('3', '5'), ('4', '5')]
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#>>>
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#>>> A = [1,1,3,3,3]
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#>>> print list(combinations(A,4))
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#[(1, 1, 3, 3), (1, 1, 3, 3), (1, 1, 3, 3), (1, 3, 3, 3), (1, 3, 3, 3)]
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from itertools import combinations_with_replacement
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st,*r = input().split()
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for j in list(combinations_with_replacement(sorted(st),int(r[0]))):
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print(*j,sep="")
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# these are the more ways to find permutations
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#>>> from itertools import combinations_with_replacement
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#>>>
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#>>> print list(combinations_with_replacement('12345',2))
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#[('1', '1'), ('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '2'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '3'), ('3', '4'), ('3', '5'), ('4', '4'), ('4', '5'), ('5', '5')]
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#>>>
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#>>> A = [1,1,3,3,3]
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#>>> print list(combinations(A,2))
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#[(1, 1), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (3, 3), (3, 3), (3, 3)]
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from itertools import permutations
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st,*r = input().split()
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for i in list(sorted(permutations(st,int(r[0])))):
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print(*i,sep="")
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# these are the more ways to find permutations
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#>>> from itertools import permutations
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#>>> print permutations(['1','2','3'])
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#<itertools.permutations object at 0x02A45210>
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#>>>
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#>>> print list(permutations(['1','2','3']))
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#[('1', '2', '3'), ('1', '3', '2'), ('2', '1', '3'), ('2', '3', '1'), ('3', '1', '2'), ('3', '2', '1')]
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#>>>
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#>>> print list(permutations(['1','2','3'],2))
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#[('1', '2'), ('1', '3'), ('2', '1'), ('2', '3'), ('3', '1'), ('3', '2')]
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#>>>
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#>>> print list(permutations('abc',3))
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#[('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'), ('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')]
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#we can also chang our for loop like: for i in list(permutations(sorted(st),int(r[0]))):

‎Itertools/itertools.product().py‎

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from itertools import product
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a= list(map(int, input().split()))
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b= list(map(int, input().split()))
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print(*(list(product(a,b))))
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# There are many ways to do product.
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#>>> from itertools import product
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#>>>
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#>>> print list(product([1,2,3],repeat = 2))
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#[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]
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#>>>
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#>>> print list(product([1,2,3],[3,4]))
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#[(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)]
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#>>>
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#>>> A = [[1,2,3],[3,4,5]]
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#>>> print list(product(*A))
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#[(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)]
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#>>>
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#>>> B = [[1,2,3],[3,4,5],[7,8]]
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#>>> print list(product(*B))
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#[(1, 3, 7), (1, 3, 8), (1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 3, 7), (2, 3, 8), (2, 4, 7), (2, 4, 8), (2, 5, 7), (2, 5, 8), (3, 3, 7), (3, 3, 8), (3, 4, 7), (3, 4, 8), (3, 5, 7), (3, 5, 8)]

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