If the reason you're checking is so you can do something like if file_exists: open_it()
, it's safer to use a try
around the attempt to open it. Checking and then opening risks the file being deleted or moved or something between when you check and when you try to open it.
If you're not planning to open the file immediately, you can use os.path.isfile
Return True
if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.
import os.path
os.path.isfile(fname)
if you need to be sure it's a file.
Starting with Python 3.4, the pathlib
module offers an object-oriented approach (backported to pathlib2
in Python 2.7):
from pathlib import Path
my_file = Path("/path/to/file")
if my_file.is_file():
# file exists
To check a directory, do:
if my_file.is_dir():
# directory exists
To check whether a Path
object exists independently of whether is it a file or directory, use exists()
:
if my_file.exists():
# path exists
You can also use resolve(strict=True)
in a try
block:
try:
my_abs_path = my_file.resolve(strict=True)
except FileNotFoundError:
# doesn't exist
else:
# exists
>>> ["foo", "bar", "baz"].index("bar")
1
Reference: Data Structures > More on Lists
Caveats follow
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
An 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)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
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)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
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.)
Best Answer
Using the implicit booleanness of the empty
list
is quite pythonic.