re-dict | redis的dict结构移植成通用模块
kandi X-RAY | re-dict Summary
kandi X-RAY | re-dict Summary
把redis的dict(字典)结构移植成通用模块. sds.* zmalloc.* dict.* fmacros.h config.h 均未做调整. 只需阅读main.c文件就可了解如何使用dict结构.
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QUESTION
I recently watched Raymond Hettingers talk about python dictionaries (and by extension sets...) and he mentioned that integers hash to themselves and that adding integers to a dict (or set...) will insert them in order and as long as you don't delete items the order will be preserved in python 3.6 (and probably above?). In the answer to this question it is stated that dictionaries preserve order of insertion, but for sets it seams like integers are ordered according to their value.
Now: according to the time-complexity section of python.org and in more detail here it is stated that the average time-complexity of adding an element to a set is O(1). This means if you have an unsorted list of integers it should be possible to sort them by simply doing:
...ANSWER
Answered 2020-Apr-08 at 10:40Generally, O(n)
sorting is possible for integers, strings and many other types of data. O(n log n) is the best you can do with sorting algorithms that only use comparisons (>
, <
, ==
) to determine the order of items, but for many types, you're not limited to such algorithms. In particular, see Radix sort for sorting integers.
QUESTION
TL;DR: How can you compare two python dictionaries if some of them have values which are unhashable/mutable (e.g. lists or pandas Dataframes)?
I have to compare dictionary pairs for equality. In that sense, this question is similar to these two, but their solutions only seem to work for immutable objects...
My problem, is that I'm dealing with pairs of highly nested dictionaries where the unhashable objects could be found in different places depending on which pair of dictionaries I'm comparing. My thinking is that I'll need to iterate across the deapest values contained in the dictionary and can't just rely on the dict.iteritems()
which only unrolls the highest key-value pairs. I'm not sure how iterate across all the possible key-value pairs contained in the dictionary and compare them either using sets/== for the hashable objects and in the cases of pandas dataframes, running df1.equals(df2).
(Note for pandas dataframe, just running df1==df2
does a piecewise comparison and NA's are poorly handled. df1.equals(df2)
gets around that does the trick.)
So for example:
...ANSWER
Answered 2017-Apr-19 at 21:33Well, there's a way to make any type comparable: Simply wrap it in a class that compares like you need it:
QUESTION
I understand dictionaries are insertion ordered in Python 3.6+, as an implementation detail in 3.6 and official in 3.7+.
Given they are ordered, it seems strange that no methods exist to retrieve the ith item of a dictionary by insertion order. The only solutions available appear to have O(n) complexity, either:
- Convert to a list via an O(n) process and then use
list.__getitem__
. enumerate
dictionary items in a loop and return the value when the desired index is reached. Again, with O(n) time complexity.
Since getting an item from a list
has O(1) complexity, is there a way to achieve the same complexity with dictionaries? Either with regular dict
or collections.OrderedDict
would work.
If it's not possible, is there a structural reason preventing such a method, or is this just a feature which has not yet been considered / implemented?
...ANSWER
Answered 2018-Sep-25 at 23:37For an OrderedDict
it's inherently O(n)
because the ordering is recorded in a linked list.
For the builtin dict, there's a vector (a contiguous array) rather than a linked list, but pretty much the same thing in the end: the vector contains a few kind of "dummies", special internal values that mean "no key has been stored here yet" or "a key used to be stored here but no longer". That makes, e.g., deleting a key extremely cheap (just overwrite the key with a dummy value).
But without adding auxiliary data structures on top of that, there's no way to skip over the dummies without marching over them one at a time. Because Python uses a form of open addressing for collision resolution, and keeps the load factor under 2/3, at least a third of the vector's entries are dummies. the_vector[i]
can be accessed in O(1)
time, but really has no predictable relation to the i'th non-dummy entry.
QUESTION
I have two json files, each in the form of a dictionary. I would like to compare them but ignore the 'ver' key when doing so. I've looked at the following question and implemented the answer as my function: Compare dictionaries ignoring specific keys
However, the function is still returning false when comparing two files that only have a difference in the 'ver' key.
...ANSWER
Answered 2018-Jul-23 at 19:24Change ('ver')
to ('ver',)
.
('ver')
is not a tuple, it is simply 'ver'
in the parentheses. Respectively, set(('ver'))
is {'e','r','v'}
, which are the keys that your function ignores - but they are not the keys that you want to ignore.
On the contrary, ('ver',)
is a one-element tuple, and set(('ver',))
is {'ver'}
.
QUESTION
I'm reading @Martijn Pieters' response to Converting dict to OrderedDict. The main point of his answer is that passing a regular dict to OrderedDict()
will not retain the order as desired, because the dict that you are passing has already "lost" any semblance of order. His solution is to pass tuples that make up the dict's key/value pairs instead.
However, I also noticed the following in the docs:
Changed in version 3.6: With the acceptance of PEP 468, order is retained for keyword arguments passed to the OrderedDict
Does this invalidate the issue that Martijn points out (can you now pass a dict to OrderedDict), or am I misinterpreting?
...ANSWER
Answered 2018-Mar-15 at 11:12Order is retained for keyword arguments passed to the OrderedDict
What this means is that the following is guaranteed to preserve the order:
QUESTION
I'm trying to scrape the example sentences for a specific french word using python, but the page I get back into python doesn't seem to have any results.
I've inspected the element of the search box and search button and included them as parameters. Perhaps I'm missing something?
...ANSWER
Answered 2017-Jul-25 at 18:39This works for me. Notice that I used the GET
method and the URI that is referenced in the actual form on that page.
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