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### Table not constructed becaus list is not iterable

Sorry but one more time I encounter a surprising problem

ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5]
cond1 = ['\tiny${}{}{}{}$'.format(*w) for w in Ap]


are two lists (I have verifies by type(). But when I type

table(ne1, header_column=cond1)


the return is an error 'sage.rings.integer.Integer' object is not iterable. There is something I miss.

### Table not constructed becaus list is not iterable

Sorry but one more time I encounter a surprising problem

ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5]
cond1 = ['\tiny${}{}{}{}$'.format(*w) for w in Ap]


are two lists (I have verifies by type(). But when I type

table(ne1, table([ne1], header_column=cond1)


\tiny doesn't work and the return is an error 'sage.rings.integer.Integer' object is cells are not iterable. There is something I miss.centered. Is it possible ?

### Table not constructed becaus list is not iterable

Sorry but one more time I encounter a surprising problem

ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5]
cond1 = ['\tiny${}{}{}{}$'.format(*w) for w in Ap]


are two lists (I have verifies by type(). But when I type

table([ne1], header_column=cond1)


\tiny doesn't work and the cells are not centered. Is it possible ?

### Table not constructed becaus list is not iterableCentering the cells of a table and the use of tiny size

Sorry but one more time I encounter a surprising problemchange my question to be clearer. In the following code (I have integrated the solution for \tiny :

ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5]
cond1 = ['\tiny${}{}{}{}$'.format(*w) ne=ne1
Ap=All_pref(cand,2)
cond1=['$\\tiny${}{}{}{}$'.format(*w) for w in Ap] cond2 =[""]+ [r'${}$'.format(*w) for w in cand] cond3 =[r'${}$'.format(*w) for w in cand] t1=table([ne],header_row=cond1) rank=[[x.find(l) for x in Ap] for l in cand] score_de_Borda=[[abs(x.find(l)-len(cand)) for x in Ap] for l in cand] t2=table(rank,header_row=cond1,header_column=cond2) total_score_de_Borda=[add([score_de_Borda[j][i]*ne[i] for i in range(24)]) for j in range(len(cand))] #t3=table(total_score_de_Borda,header_column=cond3) show(cond3,total_score_de_Borda)  are two lists (I have verifies by all works fine les the type(). But when I type table([ne1], header_column=cond1)  \tiny,header_column=cond3 doesn't work (without it I have no problem) and as show(cond3,total_score_de_Borda) shows the cells are not centered. Is it possible ?two elements have the same length. ### Centering the cells of a table and the use of tiny size I change my question to be clearer. In the following code (I have integrated the solution for \tiny : cand=["A","B","C","D"] ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5] ne=ne1 Ap=All_pref(cand,2) cond1=['$\\tiny${}{}{}{}$'.format(*w) for w in Ap]
cond2 =[""]+ [r'${}$'.format(*w) for w in cand]
cond3 =[r'${}$'.format(*w) for w in cand]
rank=[[x.find(l) for x in Ap] for l in cand]
score_de_Borda=[[abs(x.find(l)-len(cand)) for x in Ap] for l in cand]
total_score_de_Borda=[add([score_de_Borda[j][i]*ne[i] for i in range(24)]) for j in range(len(cand))]
show(cond3,total_score_de_Borda)


all works fine les the ,header_column=cond3 (without it I have no problem) and as show(cond3,total_score_de_Borda) shows the two elements have the same length.

### Centering the cells of a table and the use of tiny size

I change my question to be clearer. In the following code (I have integrated the solution for \tiny :

def All_pref(cand=["A","B","C","D"],code=1) : ncand=len(cand) Scand=sorted(Set(cand)) all_pref=Arrangements(Scand,ncand).list() all_pref1=[str(Word(x)) for x in all_pref] if code==1 : return ncand if code==2 : return all_pref1

cand=["A","B","C","D"]
ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5]
ne=ne1
Ap=All_pref(cand,2)
cond1=['$\\tiny${}{}{}{}$'.format(*w) for w in Ap] cond2 =[""]+ [r'${}$'.format(*w) for w in cand] cond3 =[r'${}$'.format(*w) for w in cand] t1=table([ne],header_row=cond1) rank=[[x.find(l) for x in Ap] for l in cand] score_de_Borda=[[abs(x.find(l)-len(cand)) for x in Ap] for l in cand] t2=table(rank,header_row=cond1,header_column=cond2) total_score_de_Borda=[add([score_de_Borda[j][i]*ne[i] for i in range(24)]) for j in range(len(cand))] #t3=table(total_score_de_Borda,header_column=cond3) show(cond3,total_score_de_Borda)  all works fine les the ,header_column=cond3 (without it I have no problem) and as show(cond3,total_score_de_Borda) shows the two elements have the same length.  7 None slelievre 14569 ●16 ●136 ●287 http://carva.org/samue... ### Centering the cells of a table and the use of tiny size I change my question to be clearer. clarify. In the following code (I have integrated the solution for \tiny : ): First define a function: def All_pref(cand=["A","B","C","D"],code=1) All_pref(cand=["A", "B", "C", "D"], code=1) : ncand=len(cand) Scand=sorted(Set(cand)) all_pref=Arrangements(Scand,ncand).list() all_pref1=[str(Word(x)) ncand = len(cand) Scand = sorted(Set(cand)) all_pref = Arrangements(Scand, ncand).list() all_pref1 = [str(Word(x)) for x in all_pref] if code==1 : code == 1: return ncand if code==2 : code == 2: return all_pref1all_pref1  Then use it: cand=["A","B","C","D"] ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5] ne=ne1 Ap=All_pref(cand,2) cond1=['$\\tiny${}{}{}{}$'.format(*w) cand = ["A", "B", "C", "D"]
ne1 = [18, 16, 14, 12, 11, 20, 19, 14, 16, 12, 2, 1, 0, 0, 20, 16, 13, 15, 11, 10, 9, 8, 7, 5]
ne = ne1
Ap = All_pref(cand,2)
cond1 = ['$\\tiny${}{}{}{}$'.format(*w) for w in Ap] cond2 =[""]+ = [""] + [r'${}$'.format(*w) for w in cand] cond3 =[r'${}$'.format(*w) = [r'${}$'.format(*w) for w in cand] t1=table([ne],header_row=cond1) rank=[[x.find(l) t1 = table([ne], header_row=cond1) rank = [[x.find(l) for x in Ap] for l in cand] score_de_Borda=[[abs(x.find(l)-len(cand)) score_de_Borda = [[abs(x.find(l) - len(cand)) for x in Ap] for l in cand] t2=table(rank,header_row=cond1,header_column=cond2) total_score_de_Borda=[add([score_de_Borda[j][i]*ne[i] t2 = table(rank,header_row=cond1,header_column=cond2) total_score_de_Borda = [add([score_de_Borda[j][i]*ne[i] for i in range(24)]) for j in range(len(cand))] #t3=table(total_score_de_Borda,header_column=cond3) # t3 = table(total_score_de_Borda,header_column=cond3) show(cond3,total_score_de_Borda)  all everything works fine les apart from the ,header_column=cond3 (without it I have no problem) and as show(cond3,total_score_de_Borda) shows show(cond3, total_score_de_Borda) shows, the two elements have the same length.  8 None slelievre 14569 ●16 ●136 ●287 http://carva.org/samue... ### Centering the cells of a table and the use of tiny size I change my question to clarify. In the following code (I have integrated the solution for \tiny): First define a function: def All_pref(cand=["A", "B", "C", "D"], code=1) : ncand = len(cand) Scand = sorted(Set(cand)) all_pref = Arrangements(Scand, ncand).list() all_pref1 = [str(Word(x)) for x in all_pref] if code == 1: return ncand if code == 2: return all_pref1  Then use it: cand = ["A", "B", "C", "D"] ne1 = [18, 16, 14, 12, 11, 20, 19, 14, 16, 12, 2, 1, 0, 0, 20, 16, 13, 15, 11, 10, 9, 8, 7, 5] ne = ne1 Ap = All_pref(cand,2) All_pref(cand, 2) cond1 = ['$\\tiny${}{}{}{}$'.format(*w) for w in Ap]
cond2 = [""] + [r'${}$'.format(*w) for w in cand]
cond3 = [r'${}$'.format(*w) for w in cand]

everything works fine apart from the ,header_column=cond3header_column=cond3 (without it I have no problem) and as show(cond3, total_score_de_Borda) shows, the two elements have the same length.