python
import sys sys.path.append("..")
import redbaron redbaron.ipython_behavior = False
from redbaron import RedBaron
As you have seen in the previous section, you can navigate into RedBaron tree only using attribute access and index access on list of nodes with the use of the .help() method to know what you can do. However, RedBaron offers way more powerful and convenient tools to do that.
To retrieve a single node, you can use the .find() method by passing it one of the identifiers listed in .help() of node you want to get, this way:
python
red = RedBaron("a = 1")
red.help()
red.find('NameNode').help() red.find('namenode').help() # identifiers are not case sensitive red.find('name')
This will recursively travel the tree and return the first node of that type.
You can also specify attributes of the node that you want to match:
In [36]: red = RedBaron("a = b")
In [37]: red.find('name').help()
In [38]: red.find('name', value='b').help()
If you don't want a recursive approach but only on the first level on the current node or node list, you can pass recursive=False to .find().
Like BeautifulSoup, RedBaron provides a shorthand to .find(), you can write the name of the target as an attribute of the node and this will do a .find() in the same fashion:
In [39]: red = RedBaron("a = b")
In [40]: red.find('name')
In [41]: red.name
You might have noticed that some identifiers end with a _, those are for the case where the identifier might be a Python reserved keyword like if, or while for example.
.find_all() is extremely similar to .find() except it returns a node list containing all the matching queries instead of a single one. Like in BeautifulSoup, __call__ is aliased to find_all (meaning that if you try to call the node this way node(some_arguments) this will call .find_all() with the arguments).
In [45]: red = RedBaron("a = b")
In [46]: red.find_all("NameNode")
In [47]: red.find_all("name")
In [48]: red.findAll("name")
In [49]: red.findAll("name", value="b")
In [50]: red("name", value="b")
.find_all() also supports the option recursive=False.
.find() and .find_all() offer more powerful comparison mean than just equality comparison.
Instead of passing a string to test properties of the identifier of a node, you can pass a callable, like a lambda. It will receive the value as first argument:
python
red = RedBaron("a = [1, 2, 3, 4]") red.find("int", value=lambda value: int(value) % 2 == 0) red.find_all("int", value=lambda value: int(value) % 2 == 0) red.find(lambda identifier: identifier == "comma") red.find_all(lambda identifier: identifier == "comma")
Instead of passing a string to test properties of a node, you can pass a compiled regex:
python
import re red = RedBaron("abcd = plop + pouf") red.find("name", value=re.compile("^p")) red.find_all("name", value=re.compile("^p")) red.find(re.compile("^n")) red.find_all(re.compile("^n"))
Having to compile regex is boring, so you can use this shorthand syntaxe instead (prefixing a string with "re:"):
python
red = RedBaron("abcd = plop + pouf") red.find("name", value="re:^p") red.find_all("name", value="re:^p") red.find("re:^n") red.find_all("re:^n")
Same than in a shell, you can use globs by prefixing the string with "g:":
python
red = RedBaron("abcd = plop + pouf") red.find("name", value="g:p*") red.find_all("name", value="g:p*") red.find("g:n*") red.find_all("g:n*")
In the background, the comparison is done using the fnmatch module of the standard lib.
You can pass a list as a shorthand to test if the tested attribute is in any of the member of the list/tuple:
python
red = RedBaron("foonbarnbaz") red.find("name", value=["foo", "baz"]) red.find("name", value=("foo", "baz")) red("name", value=["foo", "baz"]) red("name", value=("foo", "baz"))
python
red = RedBaron("1nstuffn'string'n") red.find(["int", "string"]) red(["int", "string"])
You can also pass as namy callable as args (without giving it a key) as you want, those callables will receive the node itself as first argument (and must return a value that will be tested as a boolable):
python
red = RedBaron("a = [1, 2, 3, 4]") red.find("int", lambda node: int(node.value) % 2 == 0) red.find_all("int", lambda node: int(node.value) % 2 == 0) red.find("int", lambda node: int(node.value) % 2 == 0, lambda node: int(node.value) == 4)
To ease the usage of RedBaron in ipython (and in general), you can pass any of the previous testing methods (except the lambda) as the first argument of *args, it will be tested against the default testing attribute which is the "value" attribute by default. This mean that: red.find("name", "foo") is the equivalent of red.find("name", value="foo").
If the default tested attribute is different, it will be shown in .help(). For now, the 2 only cases where this happens is on class node and funcdef node where the attribute is "name".
python
red = RedBaron("foondef bar(): passnbazndef badger(): pass") red.find("name", "baz") red.find("def", "bar") red.find("def").help()
To learn how to modify stuff in RedBaron read modifying.