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Maven dependency for Servlet 3.0 API?
How can I tell Maven 2 to load the Servlet 3.0 API?
10 Answers
10
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COUNT DISTINCT with CONDITIONS
...count(distinct tag) as tag_count,
count(distinct (case when entryId > 0 then tag end)) as positive_tag_count
from
your_table_name;
The first count(distinct...) is easy.
The second one, looks somewhat complex, is actually the same as the first one, except that you use case...when clause. In ...
Code Golf: Lasers
...
Perl, 166 160 characters
Perl, 251 248 246 222 214 208 203 201 193 190 180 176 173 170 166 --> 160 chars.
Solution had 166 strokes when this contest ended, but A. Rex has found a couple ways to shave off 6 more characters:
s!.!$t{...
Does MSTest have an equivalent to NUnit's TestCase?
... note on an older version of the TestAdapter, which was removed from the 2.0.0's description page:
Note that it doesn't work with VS Express
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surface plots in matplotlib
...t.figure()
ax = fig.add_subplot(111, projection='3d')
x = y = np.arange(-3.0, 3.0, 0.05)
X, Y = np.meshgrid(x, y)
zs = np.array(fun(np.ravel(X), np.ravel(Y)))
Z = zs.reshape(X.shape)
ax.plot_surface(X, Y, Z)
ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')
plt.show()
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Google Maps API v3: How do I dynamically change the marker icon?
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answered Dec 21 '09 at 17:01
Sudhir JonathanSudhir Jonathan
15.3k1111 gold badges5959 silver badges8585 bronze badges
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Iterating over a numpy array
...>>> for (x,y), value in numpy.ndenumerate(a):
... print x,y
...
0 0
0 1
1 0
1 1
2 0
2 1
Regarding the performance. It is a bit slower than a list comprehension.
X = np.zeros((100, 100, 100))
%timeit list([((i,j,k), X[i,j,k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k...
How to iterate a loop with index and element in Swift
...
Yes. As of Swift 3.0, if you need the index for each element along with its value, you can use the enumerated() method to iterate over the array. It returns a sequence of pairs composed of the index and the value for each item in the array. For...
Accessing nested JavaScript objects and arays by string path
... // strip a leading dot
var a = s.split('.');
for (var i = 0, n = a.length; i < n; ++i) {
var k = a[i];
if (k in o) {
o = o[k];
} else {
return;
}
}
return o;
}
Usage::
Object.byString(someObj, 'part3[0].name');
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Is a Python list guaranteed to have its elements stay in the order they are inserted in?
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answered Dec 4 '12 at 0:11
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