For Python 3.5+ use:
spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py")
foo = importlib.util.module_from_spec(spec)
For Python 3.3 and 3.4 use:
from importlib.machinery import SourceFileLoader
foo = SourceFileLoader("module.name", "/path/to/file.py").load_module()
(Although this has been deprecated in Python 3.4.)
For Python 2 use:
foo = imp.load_source('module.name', '/path/to/file.py')
There are equivalent convenience functions for compiled Python files and DLLs.
See also http://bugs.python.org/issue21436.
>>> ["foo", "bar", "baz"].index("bar")
Reference: Data Structures > More on Lists
Note that while this is perhaps the cleanest way to answer the question as asked,
index is a rather weak component of the
list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about
list.index follow. It is probably worth initially taking a look at the documentation for it:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a
ValueError if there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
Linear time-complexity in list length
index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give
index a hint. For instance, in this snippet,
l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight
l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:
>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
Only returns the index of the first match to its argument
A call to
index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.
>>> [1, 1].index(1)
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
Most places where I once would have used
index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for
index, take a look at these excellent Python features.
Throws if element not present in list
A call to
index results in a
ValueError if the item's not present.
>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
If the item might not be present in the list, you should either
- Check for it first with
item in my_list (clean, readable approach), or
- Wrap the
index call in a
try/except block which catches
ValueError (probably faster, at least when the list to search is long, and the item is usually present.)
Since you don't provide any information on how you want to format your XML, i just invented my own notation.
To print to console, iterate over items in a similar way but change the output (by the