Replacing each character in a string using an index into a list

Question

Suppose I have a string as follows:

mystr = "MY VALUES ARE: (?, ?, ?, ?)"
values = ['a', 'b', 'f', 12]

I would like replace each ? with the corresponding value at that index in the list: values. So first ? would be replaced with values[0], second ? would be replaced with values[1]

So my final str would look like:

MY VALUES ARE: ('a', 'b', 'f', '12')

NOTE: The number of ? will vary but it will always equal the number of values in values

• cs95: You're leaving yourself open to SQL injection. Don't do this.

Answer 1

You can replace the ? with {} and call format:

print(mystr.replace("?", "'{}'").format(*values))
#MY VALUES ARE: ('a', 'b', 'f', '12')

Answer 2

This isn't sql, this just looks like it. Don't worry about SQL injection, it's not a concern here.

See https://bobby-tables.com/python for parametrized queries - for simply replacement use str.replace(old, new, count=1)

sql = "My VALUES are (?, ?, ?, ?)"
values = ['a', 'b', 'f', 12]
for v in values:
 sql = sql.replace("?",f"'{v}'",1) # inefficient - will create intermediate strings to
 # be replaced
print(sql)

Output:

My VALUES are ('a', 'b', 'f', '12')

---

Slightly more performant but more code as well:

sql = "My VALUES are (?, ?, ?, ?)"
values = ['a', 'b', 'f', 12]
k = iter(values)
# see list comp below for shorter approach
l = [] 
for c in sql:
 if c != '?':
 l.append(c)
 else:
 l.append(f"'{next(k)}'")
sql = "".join(l)
print(sql) # My VALUES are ('a', 'b', 'f', '12') as well

As list comprehension (join is faster on list then on generator comps) thx @Ev. Kuonis :

sql = "".join( [ c if c != '?' else f"'{next(k)}'" for c in sql] )

• Ma0: the list comprehension version of the second snippet would not be as big.
• Ma0: Sorry for being pedantic but we have it on good authority that join is faster with list comprehensions than generators.
• Ma0: I reacted the same way when I found out :))