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How to validate an Email in PHP?
... Manual should suffice.
Update 1: As pointed out by @binaryLV:
PHP 5.3.3 and 5.2.14 had a bug related to
FILTER_VALIDATE_EMAIL, which resulted in segfault when validating
large values. Simple and safe workaround for this is using strlen()
before filter_var(). I'm not sure about 5.3.4 fin...
Is there a decorator to simply cache function return values?
...
Starting from Python 3.2 there is a built-in decorator:
@functools.lru_cache(maxsize=100, typed=False)
Decorator to wrap a function with a memoizing callable that saves up to the maxsize most recent calls. It can save time when an expensive or...
What's the function like sum() but for multiplication? product()?
...
Update:
In Python 3.8, the prod function was added to the math module. See: math.prod().
Older info: Python 3.7 and prior
The function you're looking for would be called prod() or product() but Python doesn't have that function. So, you need...
What is the good python3 equivalent for auto tuple unpacking in lambda?
...
33
No, there is no other way. You covered it all. The way to go would be to raise this issue on th...
Bootstrap: How do I identify the Bootstrap version?
...ootstrap.css you should have comments like the below:
/*!
* Bootstrap v2.3.1
*
* Copyright 2012 Twitter, Inc
* Licensed under the Apache License v2.0
* http://www.apache.org/licenses/LICENSE-2.0
*
* Designed and built with all the love in the world @twitter by @mdo and @fat.
*/
If they ar...
Joda-Time: what's the difference between Period, Interval and Duration?
...
3 classes are needed because they represent different concepts so it is a matter of picking the appropriate one for the job rather than of relative performance. From the documentation with comments added by me in italics:
...
cartesian product in pandas
...import DataFrame, merge
df1 = DataFrame({'key':[1,1], 'col1':[1,2],'col2':[3,4]})
df2 = DataFrame({'key':[1,1], 'col3':[5,6]})
merge(df1, df2,on='key')[['col1', 'col2', 'col3']]
Output:
col1 col2 col3
0 1 3 5
1 1 3 6
2 2 4 5
3 2 4 6
See here...
Flatten nested dictionaries, compressing keys
...;> flatten({'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3]})
{'a': 1, 'c_a': 2, 'c_b_x': 5, 'd': [1, 2, 3], 'c_b_y': 10}
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Add column with constant value to pandas dataframe [duplicate]
...8]: from numpy.random import randint
In [9]: df = DataFrame({'a': randint(3, size=10)})
In [10]:
In [10]: df
Out[10]:
a
0 0
1 2
2 0
3 1
4 0
5 0
6 0
7 0
8 0
9 0
In [11]: s = df.a[:5]
In [12]: dfa, sa = df.align(s, axis=0)
In [13]: dfa
Out[13]:
a
0 0
1 2
2 0
3 1
4 0
5 0
6 ...
What would cause an algorithm to have O(log n) complexity?
...o have that log2 16 = 4. Hmmm... what about 128?
128 / 2 = 64
64 / 2 = 32
32 / 2 = 16
16 / 2 = 8
8 / 2 = 4
4 / 2 = 2
2 / 2 = 1
This took seven steps, and log2 128 = 7. Is this a coincidence? Nope! There's a good reason for this. Suppose that we divide a number n by 2 i times. Then...