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Get statistics for each group (such as count, mean, etc) using pandas GroupBy?
... counts
– alvitawa
Jun 24 '19 at 16:04
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Regular expression for matching HH:MM time format
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Your original regular expression has flaws: it wouldn't match 04:00 for example.
This may work better:
^([0-1]?[0-9]|2[0-3]):[0-5][0-9]$
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Linear Regression and group by in R
...ual time series with one field for year (22 years) and another for state (50 states). I want to fit a regression for each state so that at the end I have a vector of lm responses. I can imagine doing for loop for each state then doing the regression inside the loop and adding the results of each reg...
jQuery animate backgroundColor
... g.elem.style[e] = "rgb(" + [Math.max(Math.min(parseInt((g.pos * (g.end[0] - g.start[0])) + g.start[0]), 255), 0), Math.max(Math.min(parseInt((g.pos * (g.end[1] - g.start[1])) + g.start[1]), 255), 0), Math.max(Math.min(parseInt((g.pos * (g.end[2] - g.start[2])) + g.start[2]), 255), 0)].join(",") ...
Are list-comprehensions and functional functions faster than “for loops”?
...el loop:
>>> dis.dis(<the code object for `[x for x in range(10)]`>)
1 0 BUILD_LIST 0
3 LOAD_FAST 0 (.0)
>> 6 FOR_ITER 12 (to 21)
9 STORE_FAST 1 (x)
12 LOAD_FAST...
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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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...
Convert light frequency to RGB?
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answered Sep 24 '09 at 15:47
Stephen MesaStephen Mesa
4,31333 gold badges2121 silver badges1616 bronze badges
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Normalize data in pandas
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In [92]: df
Out[92]:
a b c d
A -0.488816 0.863769 4.325608 -4.721202
B -11.937097 2.993993 -12.916784 -1.086236
C -5.569493 4.672679 -2.168464 -9.315900
D 8.892368 0.932785 4.535396 0.598124
In [93]: df_norm = (df - df.mean()) / (df.max() - df...
Fastest way to determine if an integer's square root is an integer
...ts. (I found looking at the last six didn't help.) I also answer yes for 0. (In reading the code below, note that my input is int64 x.)
if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )
return false;
if( x == 0 )
return true;
Next, check if it's a square modulo 25...