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Examples:
Input: n = 14 Output: 3 The binary representation of 14 is 111 0. Input: n = 222 Output: 4 The binary representation of 222 is 110 1111 0.
We have a solution to this problem, please refer to Length of longest sequential 1 in binary . We can fix this problem quickly in Python. The approach is very simple,
 Convert decimal to binary using bin () function returns the binary representation of a number in string form and prefixes 0b.
 Separate all substrings of consecutive 1s, separated by zeros, with using the string method split () .
 Print the maximum length of split substrings from 1.
# Function to find the length of the longest sequential
# 1 in binary
def
maxConsecutive1 (
input
):
# convert number to binary
input
=
bin
(
input
)
# remove the first two characters of the output line
input
=
input
[
2
:]
# input.split (& # 39; 0 & # 39;)  & gt; breaks all substrings
# consecutive 1s separated by 0, the output will be
# be like [& # 39; 11 & # 39 ;, & # 39; 1111 & # 39;]
# map (len, input.split (& # 39; 0 & # 39;))  & gt; map map functions
Function # len in each substring of consecutive 1
# max () returns the maximum item in the list
print
(
max
(
map
(
len
,
input
. split (
’0 ’
))))
# Driver program
if
__ name__
=
=
’__main__’
:
input
=
222
maxConsecutive1 (
input
)
Output:
4
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Python Map  Length of longest sequential 1 in binary representation of a given integer 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 zerobased 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 timecomplexity 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
We hope this article has helped you to resolve the problem. Apart from Python Map  Length of longest sequential 1 in binary representation of a given integer, check other findrelated topics.
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