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How to convert a scala.List to a java.util.List?
...ort scala.collection.jcl.ArrayList
unconvertList(new ArrayList ++ List(1,2,3))
From Scala 2.8 onwards:
import scala.collection.JavaConversions._
import scala.collection.mutable.ListBuffer
asList(ListBuffer(List(1,2,3): _*))
val x: java.util.List[Int] = ListBuffer(List(1,2,3): _*)
However, asLis...
Iterate a list with indexes in Python
...ld swear I've seen the function (or method) that takes a list, like this [3, 7, 19] and makes it into iterable list of tuples, like so: [(0,3), (1,7), (2,19)] to use it instead of:
...
Auto-center map with multiple markers in Google Maps API v3
This is what I use to display a map with 3 pins/markers:
7 Answers
7
...
Finding differences between elements of a list
...
>>> t
[1, 3, 6]
>>> [j-i for i, j in zip(t[:-1], t[1:])] # or use itertools.izip in py2k
[2, 3]
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Find element's index in pandas Series
...
>>> myseries[myseries == 7]
3 7
dtype: int64
>>> myseries[myseries == 7].index[0]
3
Though I admit that there should be a better way to do that, but this at least avoids iterating and looping through the object and moves it to the C level....
Obfuscated C Code Contest 2006. Please explain sykes2.c
...g:
main(_) {
_^448 && main(-~_);
putchar(--_%64
? 32 | -~7[__TIME__-_/8%8][">'txiZ^(~z?"-48] >> ";;;====~$::199"[_*2&8|_/64]/(_&2?1:8)%8&1
: 10);
}
Introducing variables to untangle this mess:
main(int i) {
if(i^448)
main(-~i);
...
Determining complexity for recursive functions (Big O notation)
...
362
The time complexity, in Big O notation, for each function:
int recursiveFun1(int n)
{
if ...
Pickle incompatibility of numpy arrays between Python 2 and 3
I am trying to load the MNIST dataset linked here in Python 3.2 using this program:
7 Answers
...
No secret option provided to Rack::Session::Cookie warning?
I am running Rails 3.2.3, Ruby 1.9 under Fedora 17. I get this warning, when I run rails s , and how do I fix?
7 Answers
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Why are Python lambdas useful? [closed]
...where you can pass functions to other functions to do stuff. Example:
mult3 = filter(lambda x: x % 3 == 0, [1, 2, 3, 4, 5, 6, 7, 8, 9])
sets mult3 to [3, 6, 9], those elements of the original list that are multiples of 3. This is shorter (and, one could argue, clearer) than
def filterfunc(x):
...