大约有 740 项符合查询结果（耗时：0.0108秒） [XML]
Java time-based map/cache with expiring keys [closed]
...()
.concurrencyLevel(4)
.softKeys()
.weakValues()
.maximumSize(10000)
.expiration(10, TimeUnit.MINUTES)
.makeComputingMap(
new Function<Key, Graph>() {
public Graph apply(Key key) {
return createExpensiveGraph(key);
}
});
Update:
...
How to add parameters to HttpURLConnection using POST using NameValuePair
...tion conn = (HttpsURLConnection) url.openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair(...
Remove all occurrences of a value from a list?
...
Consider x = [1] * 10000 + [2] * 1000. The loop body executes 1000 times and .remove() has to skip 10000 elements every time it's invoked. That smells like O(n*n) to me but is no proof. I think the proof would be to assume that the number of 2s...
Using numpy to build an array of all combinations of two arrays
...
@pv's solution
In [113]:
%timeit cartesian(([1, 2, 3], [4, 5], [6, 7]))
10000 loops, best of 3: 135 µs per loop
In [114]:
cartesian(([1, 2, 3], [4, 5], [6, 7]))
Out[114]:
array([[1, 4, 6],
[1, 4, 7],
[1, 5, 6],
[1, 5, 7],
[2, 4, 6],
[2, 4, 7],
[2, 5, 6...
How to Sign an Already Compiled Apk
...release-key.keystore -alias alias_name -keyalg RSA -keysize 2048 -validity 10000
then sign the apk using :
jarsigner -verbose -sigalg SHA1withRSA -digestalg SHA1 -keystore my-release-key.keystore my_application.apk alias_name
check here for more info
...
UnicodeDecodeError: 'ascii' codec can't decode byte 0xd1 in position 2: ordinal not in range(128)
....
Also, you can try finding the encoding automatically by reading the top 10000 bytes using the below snippet:
import chardet
with open("dataset_path", 'rb') as rawdata:
result = chardet.detect(rawdata.read(10000))
print(result)
...
Insert an element at a specific index in a list and return the updated list
...python3 -m timeit -s "a = list(range(1000))" "b = a[:]; b[500:500] = [3]"
100000 loops, best of 3: 3.08 µsec per loop
ATOzTOA's accepted answer based on merge of sliced lists - Second (6.71 µsec per loop)
mquadri$ python3 -m timeit -s "a = list(range(1000))" "b = a[:500] + [3] + a[500:]"
1000...
ExecutorService that interrupts tasks after a timeout
...ble(){
public void run(){
handler.cancel();
}
}, 10000, TimeUnit.MILLISECONDS);
This will execute your handler (main functionality to be interrupted) for 10 seconds, then will cancel (i.e. interrupt) that specific task.
...
How to find third or nth maximum salary from salary table?
...order , it has to be done in DESC order , if we have salary like this 7000,10000,11000,500,800,900,12000 , then inner query of sorting will result in top3 that means 500,800,900 and max of these is 900, but 900 is not the 3 maximum , 3 maximum salary is 10000.
– Narendra Jaggi
...
How to access the ith column of a NumPy multidimensional array?
...ave a very big array A instead of the arr:
A = np.random.randint(2, size=(10000,10000), dtype='int32')
A_c1_ref = A[:, 1]
A_c1_copy = A[:, 1].copy()
and you want to compute the sum of all the elements of the first column, i.e. A_c1_ref.sum() or A_c1_copy.sum(). Using the copied version is much f...