Extending the Python Template strings

Question

Is it possible to extent the Template strings so I can add methods to the identifier:

By default I can replace simple variables:

from string import Template
t = Template("${noun}ification")
t.substitute(dict(noun='Egg'))  # Output: 'Eggification'

But I wish to extend the identifier with a custom method (e.g. as_b):

t = Template("${noun.as_b}ification")  # Desired output: '<b>Egg</b>ification'

I know I can easily do something similar with Formatted string literals but is it possible using the Template strings, like above? I cannot use Formatted string literals because it will not work inside a script, a JSON file or whatever text that contain curly braces.

Essentially I'm looking for something that can offer the Template strings syntax within the Formatted string literals flexibility.

Answer 1

Genshi does exactly that but it doesn't seem to use the built-in string module.
Other useful references are Chameleon and Mako.

Answer 2

Not a direct answer, but I'd recommend not using Template strings.

If you're using Python 3.6 or newer, use f-strings:

def bold(t):
    return f"<b>{t}</b>"

noun = 'Egg'

print(f"{bold(noun)}ification")

when run:

<b>Egg</b>ification

If you'd rather a more powerful template mechanism, I highly recommend Jinja2.

Answer 3

Wanted to do this myself, figured it out so came to update my google hit.

from string import Template
class MyTemplate(Template):
    idpattern = r'(?a:[_a-z][_.a-z0-9]*)'
    
s = MyTemplate("$myvar $my.var and ${myvar} and ${my.var}")
s.safe_substitute({'myvar': 'aaaa', 'my.var': 'bbbb'})

Result:

'aaaa bbbb and aaaa and bbbb'

Other over-ride options are both the delimiter ($) or braceidpattern (None) or pattern:

pattern = fr"""
{delim}(?:
  (?P<escaped>{delim})  |   # Escape sequence of two delimiters
  (?P<named>{id})       |   # delimiter and a Python identifier
  {{(?P<braced>{bid})}} |   # delimiter and a braced identifier
  (?P<invalid>)             # Other ill-formed delimiter exprs
)
"""