To check if o
is an instance of str
or any subclass of str
, use isinstance (this would be the "canonical" way):
if isinstance(o, str):
To check if the type of o
is exactly str
(exclude subclasses):
if type(o) is str:
The following also works, and can be useful in some cases:
if issubclass(type(o), str):
See Built-in Functions in the Python Library Reference for relevant information.
One more note: in this case, if you're using Python 2, you may actually want to use:
if isinstance(o, basestring):
because this will also catch Unicode strings (unicode
is not a subclass of str
; both str
and unicode
are subclasses of basestring
). Note that basestring
no longer exists in Python 3, where there's a strict separation of strings (str
) and binary data (bytes
).
Alternatively, isinstance
accepts a tuple of classes. This will return True
if o
is an instance of any subclass of any of (str, unicode)
:
if isinstance(o, (str, unicode)):
>>> ["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
The magic number comes from UNIX-type systems where the first few bytes of a file held a marker indicating the file type.
Python puts a similar marker into its
pyc
files when it creates them.Then the python interpreter makes sure this number is correct when loading it.
Anything that damages this magic number will cause your problem. This includes editing the
pyc
file or trying to run apyc
from a different version of python (usually later) than your interpreter.If they are your
pyc
files, just delete them and let the interpreter re-compile thepy
files. On UNIX type systems, that could be something as simple as:or:
If they are not yours, you'll have to either get the
py
files for re-compilation, or an interpreter that can run thepyc
files with that particular magic value.One thing that might be causing the intermittent nature. The
pyc
that's causing the problem may only be imported under certain conditions. It's highly unlikely it would import sometimes. You should check the actual full stack trace when the import fails?As an aside, the first word of all my
2.5.1(r251:54863)
pyc
files is62131
,2.6.1(r261:67517)
is62161
. The list of all magic numbers can be found inPython/import.c
, reproduced here for completeness (current as at the time the answer was posted, it may have changed since then):