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Convert tuple to list and back
...confused.
– Jimmy
Aug 28 '18 at 19:30
6
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Is there a simple way to remove multiple spaces in a string?
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Francisco Couzo
8,04633 gold badges2929 silver badges3737 bronze badges
answered Oct 9 '09 at 21:52
Josh LeeJosh Lee
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C#位运算符（C#按位与、按位或 等） - 更多技术 - 清泛网 - 专注C/C++及内核技术
...算按位对运算对象的值进行非运算，即：如果某一位等于0，就将其转变为1；如果某一位等于1，就将其转变为0。
比如，对二进制的10010001进行位逻辑非运算，结果等于01101110，用十进制表示就是：
~145等于110；对二进制的01010101...
What is the easiest way to remove the first character from a string?
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I kind of favor using something like:
asdf = "[12,23,987,43"
asdf[0] = ''
p asdf
# >> "12,23,987,43"
I'm always looking for the fastest and most readable way of doing things:
require 'benchmark'
N = 1_000_000
puts RUBY_VERSION
STR = "[12,23,987,43"
Benchmark.bm(7) do |b|
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How to compare software version number using js? (only number)
...at if we have version strings that are not made up of just digits (e.g. "1.0a")?
What should happen if one version string has more parts than the other? Most likely "1.0" should be considered less than "1.0.1", but what about "1.0.0"?
Here's the code for an implementation that you can use directly...
Use CSS3 transitions with gradient backgrounds
...been some browser releases that supported transitions on gradients (e.g IE10. I tested gradient transitions in 2016 in IE and they seemed to work at the time, but my test code no longer works.)
Update: October 2018
Gradient transitions with un-prefixed new syntax [e.g. radial-gradient(...)]now conf...
Difference between break and continue statement
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answered Jan 20 '09 at 18:02
Xn0vv3rXn0vv3r
16.4k1313 gold badges5353 silver badges6363 bronze badges
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How to randomize (or permute) a dataframe rowwise and columnwise?
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Given the R data.frame:
> df1
a b c
1 1 1 0
2 1 0 0
3 0 1 0
4 0 0 0
Shuffle row-wise:
> df2 <- df1[sample(nrow(df1)),]
> df2
a b c
3 0 1 0
4 0 0 0
2 1 0 0
1 1 1 0
By default sample() randomly reorders the elements passed as the first argument. This m...
pandas dataframe columns scaling with sklearn
...;> scaler = MinMaxScaler()
>>> dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],
'B':[103.02,107.26,110.35,114.23,114.68],
'C':['big','small','big','small','small']})
>>> dfTest[['A', 'B']] = scaler.fit_transform(...
Adding a column to a data.frame
...p <- as.factor(your.df$group)
no h_freq h_freqsq group
1 1 0.40998238 0.06463876 1
2 2 0.98086928 0.33093795 1
3 3 0.28908651 0.74077119 1
4 4 0.10476768 0.56784786 1
5 1 0.75478995 0.60479945 2
6 2 0.26974011 0.95231761 2
7 3 0.53676266 0.74370154...