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Finding differences between elements of a list
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>>> 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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How do I change Bootstrap 3 column order on mobile layout?
...ith a top fixed navbar. Underneath I have two columns, one for a sidebar (3), and one for content (9). Which on desktop looks like this
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Bootstrap 3 offset on right not left
In regards to BS 3 if I wanted just a narrow column of content on the right I might use an offset class of 9 and a column of 3.
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Difference between break and continue statement
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537
break leaves a loop, continue jumps to the next iteration.
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Returning the product of a list
...rt numpy as np
import numexpr as ne
# from functools import reduce # python3 compatibility
a = range(1, 101)
%timeit reduce(lambda x, y: x * y, a) # (1)
%timeit reduce(mul, a) # (2)
%timeit np.prod(a) # (3)
%timeit ne.evaluate("prod(a)") # (4)
In t...
Remove the last three characters from a string
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read last 3 characters from string [Initially asked question]
You can use string.Substring and give it the starting index and it will get the substring starting from given index till end.
myString.Substring(myString.Length-3)
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How to format a phone number with jQuery
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Simple: http://jsfiddle.net/Xxk3F/3/
$('.phone').text(function(i, text) {
return text.replace(/(\d{3})(\d{3})(\d{4})/, '$1-$2-$3');
});
Or: http://jsfiddle.net/Xxk3F/1/
$('.phone').text(function(i, text) {
return text.replace(/(\d\d\d)(\d\d\d)(...
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):
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Simultaneously merge multiple data.frames in a list
...question was marked as a duplicate of this one so I answer here, using the 3 sample data frames below:
x <- data.frame(i = c("a","b","c"), j = 1:3, stringsAsFactors=FALSE)
y <- data.frame(i = c("b","c","d"), k = 4:6, stringsAsFactors=FALSE)
z <- data.frame(i = c("c","d","a"), l = 7:9, stri...
What's the best way to send a signal to all members of a process group?
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