Index.summary()
Pandas Index.summary()
returns the summary view of the index. This function is similar to what we have for data frames.
Syntax: Index.summary (name = None)
Returns: Summary
Example # 1: Use Index.summary ()
to find the index summary. P >
|
|
Exit:
As we can see from the output, the function returned a general index summary. p>
Example # 2: Use Index.summary () c ode> to summarize the index.
|
Output:
p>
Now we will find the index summary.
|
Exit:
As we can see in the output, the function returned a summary for the index. p >
Python | Pandas Index.summary () find: Questions
Finding the index of an item in a list
5 answers
Given a list ["foo", "bar", "baz"]
and an item in the list "bar"
, how do I get its index (1
) in Python?
Answer #1
>>> ["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 atry/except
block which catchesValueError
(probably faster, at least when the list to search is long, and the item is usually present.)
Answer #2
One thing that is really helpful in learning Python is to use the interactive help function:
>>> help(["foo", "bar", "baz"])
Help on list object:
class list(object)
...
|
| index(...)
| L.index(value, [start, [stop]]) -> integer -- return first index of value
|
which will often lead you to the method you are looking for.
Answer #3
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate()
:
for i, j in enumerate(["foo", "bar", "baz"]):
if j == "bar":
print(i)
The index()
function only returns the first occurrence, while enumerate()
returns all occurrences.
As a list comprehension:
[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
Here"s also another small solution with itertools.count()
(which is pretty much the same approach as enumerate):
from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
This is more efficient for larger lists than using enumerate()
:
$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop
Python | Pandas Index.summary () log: Questions
Python"s equivalent of && (logical-and) in an if-statement
5 answers
Here"s my code:
def front_back(a, b):
# +++your code here+++
if len(a) % 2 == 0 && len(b) % 2 == 0:
return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):]
else:
#todo! Not yet done. :P
return
I"m getting an error in the IF conditional.
What am I doing wrong?
Answer #1
You would want and
instead of &&
.
Answer #2
Python uses and
and or
conditionals.
i.e.
if foo == "abc" and bar == "bac" or zoo == "123":
# do something
How do you get the logical xor of two variables in Python?
5 answers
How do you get the logical xor of two variables in Python?
For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):
str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
print "ok"
else:
print "bad"
The ^
operator seems to be bitwise, and not defined on all objects:
>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: "str" and "str"
Answer #1
If you"re already normalizing the inputs to booleans, then != is xor.
bool(a) != bool(b)
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