我的代码需要2个小时。如何使其更快

Question

我想查找两个数据帧中是否存在重叠值。我的代码大约需要两个小时。如何使其更快。

条件： 如果list2'chr'（chr1）]等于列表'chr'（1）] 并且如果list2'start'（100028246）大于或等于list1'positoin'（100028246）小于或等于list2'end'（100029244） 输出：list1'snp'（rs75631842）list1'chr'（1）list'positon'（100029244）'是' 其他输出：list1'chr'list1'chr'list'positon''否'

list2<tab separated>
chr10   100027038   100027943
chr1    100028246   100029244

list1<tab separate>
snp chr position
rs6604985   1   1552755
rs75631842  1   100028246

Output
rs6604985        1       1552755 No
rs75631842       1       100028246 Yes

def snp_enhancer_overlap_region(self,H3K27ac_file,snp_list):
    H3K27ac_marks = pd.read_csv(self.H3K27ac_file, sep="\t",header=None,names=['chr','start','end'])
    snp_list = pd.read_csv(self.snp_list, sep="\t",header=0)

   H3K27ac_marks.loc[H3K27ac_marks['start'] > H3K27ac_marks['end'], ['start','end']] = H3K27ac_marks.loc[H3K27ac_marks['start'] > H3K27ac_marks['end'], ['end','start']].values

   for index, row in snp_list.iterrows():
        count = 0
        for index, row1 in H3K27ac_marks.iterrows():
            if 'chr'+str(row['chr']) == row1['chr']:
                if(row1['start'] <= row['position'] <= row['position']):
                    count = count +1
                else:
                    count = 0
        if count > 0:
            print(row['snp'],"\t",row['chr'],"\t",row['position'],'Yes')
        else:
            print(row['snp'],"\t",row['chr'],"\t",row['position'],'No')

enhancer_overlap = Enhancer_overlap()
data = enhancer_overlap.snp_enhancer_overlap_region('list2', 'list1')