How to replace multiple characters with replace function

Question

I want to replace multiple specific character with the function replace for example :

INPUT :(c#d)&(a#b)&(k>m)

output : should be #(c,d)&#(a,b)&(k>m)

but it work only for one input #(c,d)&(a#b)&(k#m)

this is my code

inr=str(input('entrer : '))    

inr = inr.replace('(a#b)','#(a,b)') or inr.replace('(c#d)','#(c,d)')
print(inr)

score 2 · Answer 1 · answered Dec 19 '18 at 20:06

2

You can use replace twice:

inr.replace('#',',').replace('(','#(')

#(c,d)&#(a,b)&#(k,m)

answered Dec 19 '18 at 20:06

YOLO

how about if the formula is (a>b)&(c#d)&(k|r) – Dec 19 '18 at 20:21
you want the same change? – YOLO Dec 19 '18 at 20:24
Yes the same change but the out put should be specific only for (c#d) ==> #(c,d) not for all formula – Dec 19 '18 at 20:28
if you run the same code, you get undesired result? – YOLO Dec 19 '18 at 20:28
Yes the code is applied to all the formula – Dec 19 '18 at 20:30
I want to change only the formulas (a # b) and not the others (a> b) .. – Dec 19 '18 at 20:31
i think the current code works. i dont understand where is it failing – YOLO Dec 19 '18 at 20:34
the current code return (a>b)&(c#d) ==> #(a>b)(c,d) it should be (a>b)(c,d) – Dec 19 '18 at 20:36
try: `inr[0] + inr[1:].replace(r'#',',') ` – YOLO Dec 19 '18 at 20:38
now it return (a,b) it must be #(a,b) ... – Dec 19 '18 at 20:45
i fix it with inr = inr.replace('(a#b)', '#(a,b)').replace('(c#d)','#(c,d)') but it work on a specific proposition 'a' and 'b' and 'c' and 'd' – Dec 19 '18 at 20:47

1 Answers1