How to use julia string macros with a variable?

Question

Apologies if this has already been answered, but I'm having a surprisingly hard time using the "py" string macro included with the PyCall library when I feed it a variable representing a string instead of a simple string.

Examples:

@py_str "2 + 2"

returns 4

z = "2 + 2"
@py_str z

causes a error that interpolate_pycode does not take an Expr argument, as does @py_str ($(z)).

How can I pass @py_str a string variable?

(Just to clarify - above was a toy example, I'm using it for an application where it really is necessary).

Isn't it just py""" """ (i.e. Without the _str)?

daycaster
– daycaster

2017-07-30 22:47:38 +00:00
Commented Jul 30, 2017 at 22:47 — daycaster
– daycaster, Commented Jul 30, 2017 at 22:47

Dan Getz · Accepted Answer · 2017-07-30 23:31:19Z

2

@py_str "\$\$z" is your friend (looked at macro help using ?@py_str in REPL). Better to write the same macro as py"$$z"

From the REPL help (bolded relevant part):

py".....python code....."

Evaluate the given Python code string in the main Python module.

If the string is a single line (no newlines), then the Python expression is evaluated and the result is returned. If the string is
multiple lines (contains a newline), then the Python code is compiled and evaluated in the main Python module and nothing is returned.

If the o option is appended to the string, as in py"..."o, then the return value is an unconverted PyObject; otherwise, it is
automatically converted to a native Julia type if possible.

Any $var or $(expr) expressions that appear in the Python code (except in comments or string literals) are evaluated in Julia and passed to Python via auto-generated global variables. This allows you to "interpolate" Julia values into Python code.

Similarly, ny $$var or $$(expr) expressions in the Python code are evaluated in Julia, converted to strings via string, and are pasted
into the Python code. This allows you to evaluate code where the code itself is generated by a Julia expression.

PS the little typo ny instead of any is in the package source.

edited Jul 30, 2017 at 23:31

answered Jul 30, 2017 at 23:25

Dan Getz

18.3k2 gold badges25 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

lightning Over a year ago

Thank you so much!

lightning Over a year ago

Now I'm still having trouble with using the PyCall macro using variables. When I try the following `py"from math import sin print(sin(2))"

lightning Over a year ago

Now I'm having trouble with calling py_str on strings representing multiline python programs, and I believe the issue has something to do with the way julia/python represent the newline character. 'When I call 'py"y = 2 + 2(newline) print(y)"` it prints 4; when I define z = "y = 2 + 2(newline) print(y)" and then call py"$$z" I get a syntax error. Any suggestions?

Dan Getz Over a year ago

There is a pyimport function in PyCall package. Probably it could help with the from math.... Also, the documentation of PyCall is quite clear, and is at: PyCall docs. In fact the first example does @pyimport math and uses sin.

Collectives™ on Stack Overflow

How to use julia string macros with a variable?

1 Answer 1

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related