O - group of several related algorithm-CodePudding

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Question:
Length of a string array (not necessarily)
Such as: [' s1 'and s2', 'the s3]

Enter a number of clusters (number of clusters is dynamic, but must be less than the above array length) :
For example: 2

Want results:
S1, S2, S3
- S3 S1 and S2
S2 - S3, S1

Poor name all possible combinations, of course, there can be repeated,
Also note that S1 and S2 S2 - S1 is so repetitive

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CodePudding user response:

 the from itertools import combinations 

Num=int (input (" & gt;" )) 

P=[' s1 'and s2', 'the s3', 's4') 
X=list (combinations (p, num)) 
Ps=set (p) 
Gg=[[' - & gt; '. Join (I)] + list (ps - set (I)) for I in x] 
Print (gg)

CodePudding user response:

reference 1st floor chuifengde response:

 the from itertools import combinations 

Num=int (input (" & gt;" )) 

P=[' s1 'and s2', 'the s3', 's4') 
X=list (combinations (p, num)) 
Ps=set (p) 
Gg=[[' - & gt; '. Join (I)] + list (ps - set (I)) for I in x] 
Print (gg)

Sorry brother, may be I describe not clear
To give you that array P group, for example, number of clusters is 2 occasions
I want the result
S1, S2, S3, S4
- S3 S1 and S2 - S4
- S4 S1 and S2 - S3

If the length of the array model in addition to the number of clusters is not zero, the last the rest of the remainder of the final group

If the array length is 5: [' s1 'and s2', 'the s3', 's4', 's5]
Number of clusters is 2 occasions
Should be divided into
S1, s2, s3, s4, s5
.
Number of clusters for the occasion of 3
Should be divided into:
S1, s2, s3, s4, s5
S1 and s2 - s4, s3 - s5
.

CodePudding user response:

 
The import itertools 

Lista=[' s1 'and s2, s3, s4, s5,' s6 ', 's7', 'the s8', 'integrated'] 
# lista=[' s1 'and s2', 'the s3] 
Print (len (lista)) 
Partnum=int (input (' number of clusters')) 
Rang1=[I for I in range (1, len (lista) + 2 - partnum)] 
# built a sequential array print (rang1) 
Mylist=list (itertools permutations (rang1 partnum)) 
# print (mylist) 
Mylist=list (filter (lambda x: sum (x)==len (lista), mylist)) 

Newlist=[] 
For mylist1 mylist in: 
Mylist1=list (mylist1) 
Mylist1. Sort () 
If mylist1 not in newlist: 
Newlist. Append (mylist1) 
Print (newlist) 



Def get_chuan (fenzu1, start, temp, lista, num) : 
If the start==num: 
Return temp 
# print (start, temp, num) 
Fenzu2=fenzu1 [start] 
Temp1=[] 
For I in itertools.com binations (lista, fenzu2) : 
Temp1. Append (' - '. Join (I)) 
Lista1=lista. Copy () 
For i1 in I: 
Lista1. Remove (i1) 
Start1=start + 1 
# temp. Append (temp1) 
Temp2=get_chuan (fenzu1 start1, temp1, lista1, num) 
Temp. Append (temp2) 
Return temp 

Chuan=[] 
For fenzu1 in newlist: # fenzu1=,2,6 [1] 
Temp=[] 
Num=len (fenzu1) 
Temp=get_chuan (fenzu1, start=0, temp=temp, lista=lista, num=num) 
Print (temp)

CodePudding user response:

Don't know if this meaning,

 

The from itertools import combinations 

Def get_comb (set_temp comb_num) : 
Comb_items=list (combinations (set_temp comb_num)) 
If len (set_temp) - comb_num & lt;=comb_num: 
Return the [[' - '. Join (comb_item), '-'. Join (list (set_temp - set (comb_item)))] for comb_item in comb_items] 
The else: 
Comb_items_temp=[] 
For comb_item comb_items in: 
Comb_items_next=get_comb (set_temp - set (comb_item), comb_num) 
For comb_item_next comb_items_next in: 
Temp=[' - '. Join (comb_item)] 
Temp. The extend (comb_item_next) 
Comb_items_temp. Append (temp) 
Return comb_items_temp 


Test_set={s1, s2, s3, s4, s5 '} 

Test_comb=get_comb (test_set, 2) 
Print (len (test_comb), test_comb)

CodePudding user response:

 the from itertools import permutations 
The from more_itertools import chunked 

Num=int (input (" & gt;" )) 

P=[' s1 'and s2', 'the s3', 's4', 's5', 's6'] 
L=[] 
For I in the list (permutations (p, len (p))) : 
Tmp_x=chunked (I, num) 
Tmp_y=[sorted (j) for j in tmp_x] 
Tmp_y=sorted (tmp_y, key=lambda t: (- len (t), t [0])) 
If tmp_y not in l: 
L.A. ppend (tmp_y) 

Result=[] 
For t in l: 
Result. Append ([r [0] if len (r)==1 else '- & gt;'. Join in t (r) for r]) 

For I in result: 
Print (I) 
Print (len (l))

CodePudding user response:

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