我正在做的是从.txt文件中加载URL列表，该文件应能正常工作：
def load_urls():
try:
input_file = open("filters\\inputLinks.txt", "r")
for each_line in input_file:
link = each_line.rstrip('\n')
identify_platform(link)
except Exception as e:
print("Loading URLs error: ", e)
def identify_platform(link):
try:
SEARCH_FOR = ["/node/", "/itemlist/"]
if any(found in link for found in SEARCH_FOR):
print(link)
except Exception as e:
print("Identifying URLs error: ", e)
if __name__ == "__main__":
load_urls()
然后它将检查URL是否包含SEARCH_FOR数组元素之一。如果是这样，我们将其打印到屏幕上。是否有可能使用类似以下的命令打印出找到的数组元素之一：
print(element_found + "|" + link)
答案 0 :(得分：2)
如果将您分成多行，则可能会使其更具可读性：
for found in SEARCH_FOR:
if found in link:
print(found + '|' + link)
答案 1 :(得分：1)
仅使用列表推导即可代替使用#include <intrin.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include <vector>
#include <assert.h>
#include <omp.h>
using namespace std;
void convolution(float* output, float* input, float* filter, int width, int height, int r)
{
assert(output!=NULL && input!=NULL && filter!=NULL && width>0 && height>0 && r>0);
int w1=width-1;
int h1=height-1;
int fwidth=2*r+1;
int i, j, di, dj, ii, jj;
float sum;
for (i=0;i<height;++i)
{
for (j=0;j<width;++j)
{
sum=0;
for (di=-r;di<=r;++di)
{
ii=i+di;
ii=max(min(ii,h1),0);
for (dj=-r;dj<=r;++dj)
{
jj=j+dj;
jj=max(min(jj,w1),0);
sum+=filter[dj+r+(di+r)*fwidth]*input[jj+ii*width];
}
}
output[j+i*width]=sum;
}
}
}
int main()
{
// load the image
int width=1920; // width of the image
int height=1080; // height of the image
int len=width*height; // pixels in the image
int i, j, ii, jj, i2;
float* data=(float*)malloc(sizeof(float)*len); // buffer to load the image
float* output=(float*)malloc(sizeof(float)*len); // output buffer
FILE* fp=fopen("../image.dat", "rb"); // open the image, assume that the bld directory is a subdirectory to the src directory
fread(data, sizeof(float), width*height, fp); // load the float values, the image is gray.
fclose(fp);
// set the filter
int radius=3; // filter radius
float sigma=(float)(radius/3.0); // standard deviation of the Gaussian filter
float beta=(float)(-0.5/(sigma*sigma)); // coefficient exp(beta*x*x)
int fwidth=2*radius+1; // width of the filter
int flen=fwidth*fwidth; // number of elements in the filter
float* filter=(float*)malloc(sizeof(float)*flen); // filter buffer
float sum_weight=0; // we want to normalize the filter weights
for (i=-radius;i<=radius;++i)
{
ii=(i+radius)*fwidth;
i2=i*i;
for (j=-radius;j<=radius;++j)
{
jj=j+radius+ii;
filter[jj]=exp(beta*(i2+j*j));
sum_weight+=filter[jj];
}
}
sum_weight=(float)(1.0/sum_weight);
for (i=0;i<flen;++i)
filter[i]*=sum_weight; // now the weights are normalized to sum to 1
clock_t start=clock();
convolution(output, data, filter, width, height, radius);
clock_t finish=clock();
double duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf( "convolution naive: %2.3f seconds\n", duration );
float* output2=(float*)malloc(sizeof(float)*len); // output buffer
start=clock();
your_convolution(output2, data, filter, width, height, radius);
finish=clock();
double duration2 = (double)(finish - start) / CLOCKS_PER_SEC;
printf( "your convolution: %2.3f seconds\n", duration2 );
double sum=0;
for (i=0;i<len;++i)
sum+=fabs(output[i]-output2[i]);
printf("difference of the outputs=%lf\n", sum);
printf( "The performance of your convolve is %2.1f times higher than convolution naive.\n", duration/duration2);
free(data);
free(filter);
free(output);
return 0;
}
运算符。请注意，如果两个搜索项都在目标中，这可能会给您2条结果。
any