How to replace string that contains curly bracket in Python?

Question

Please see below code to understand my problem:

str_line = data['line']  # data['line'] holds "I was born in {year}"
str_year = '1987'

Expected output: I was born in 1987

print(str_line)

I have tried using following methods but the output always the original str_line (i.e., I was born in {year}).

str_line = str_line.replace('{{year}}',str_year)

and

str_line = str_line.replace('/{year/}',str_year)

You have only single braces, and you try to replace a string within double braces. Use `str_line = str_line.replace('{year}',str_year)` — Wiktor Stribiżew, Oct 23 '20 at 16:27
if you are using user input for str_line, use string.replace, but If you just want to make code easier to read, use an f-string `str_line = f"I was born in {str_year}"` — Impossible Reality, Oct 23 '20 at 16:31
Does this answer your question? [How do I put a variable inside a string?](https://stackoverflow.com/questions/2960772/how-do-i-put-a-variable-inside-a-string) — mkrieger1, Oct 23 '20 at 16:35
Thanks @mkrieger1 , it's solved using str_line.format(year=str_year) — Arif, Oct 24 '20 at 03:13

score 1 · Accepted Answer · answered May 21 '22 at 10:43

You can use str.format() with year argument set to str_year:

str_line = "I was born in {year}"
str_year = '1987'
print( str_line.format(year=str_year) )
# => I was born in 1987

See the Python demo online

Documentation says

str.format(*args, **kwargs)
Perform a string formatting operation. The string on which this method is called can contain literal text or replacement fields delimited by braces {}. Each replacement field contains either the numeric index of a positional argument, or the name of a keyword argument. Returns a copy of the string where each replacement field is replaced with the string value of the corresponding argument.

How to replace string that contains curly bracket in Python?

1 Answers1