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Find the most frequent number in a numpy vector
...s:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html
and then probably use np.argmax:
a = np.array([1,2,3,1,2,1,1,1,3,2,2,1])
counts = np.bincount(a)
print(np.argmax(counts))
For a more complicated list (that perhaps contains negative numbers or non-integer values), you can us...
Which is faster: while(1) or while(2)?
...
jmp .L2
.seh_endproc
.ident "GCC: (tdm64-2) 4.8.1"
With -O2 and -O3 (same output):
.file "main.c"
.intel_syntax noprefix
.def __main; .scl 2; .type 32; .endef
.section .text.startup,"x"
.p2align 4,,15
.globl main
.def main; .scl 2; ....
逆向工程——二进制炸弹(CSAPP Project) - 操作系统(内核) - 清泛网 - 专注...
...x8(%ebp)输入字符串首字符地址，%ecx = -0xb(%ebp)
另外，根据and $0xf, %eax和mov 0x804a5c0(%eax), %al这两条指令可以知道，source->dest的赋值是根据我们输入每个字符串各位的最低四位(%eax)+0x804a5c0中地址所对应的字符回传给dest的对应地址。...
Javascript - sort array based on another array
Is it possible to sort and rearrange an array that looks like this:
21 Answers
21
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How to get a random number in Ruby
How do I generate a random number between 0 and n ?
17 Answers
17
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How can I create a correlation matrix in R?
...correlation. So, the number of rows must be the same between your matrix x and matrix y. Ex.:
set.seed(1)
x <- matrix(rnorm(20), nrow=5, ncol=4)
y <- matrix(rnorm(15), nrow=5, ncol=3)
COR <- cor(x,y)
COR
image(x=seq(dim(x)[2]), y=seq(dim(y)[2]), z=COR, xlab="x column", ylab="y column")
text...
Pretty printing XML in Python
...ered Jul 30 '09 at 14:12
Ben NolandBen Noland
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推荐系统算法初探 - 更多技术 - 清泛网 - 专注C/C++及内核技术
...要买啥。在经济学中，有一个著名理论叫长尾理论（The Long Tail）。
套用在互联网领域中，指的就是最热的那一小部分资源将得到绝大部分的关注，而剩下的很大一部分资源却鲜少有人问津。这不仅造成了资源利用上的浪费，...
Replace non-ASCII characters with a single space
...ad:
return ''.join([i if ord(i) < 128 else ' ' for i in text])
This handles characters one by one and would still use one space per character replaced.
Your regular expression should just replace consecutive non-ASCII characters with a space:
re.sub(r'[^\x00-\x7F]+',' ', text)
Note the + t...
How to display long messages in logcat
...1000 then you can split the string you want to log with String.subString() and log it in pieces. For example:
int maxLogSize = 1000;
for(int i = 0; i <= veryLongString.length() / maxLogSize; i++) {
int start = i * maxLogSize;
int end = (i+1) * maxLogSize;
end = end > veryLongStri...