我在文件中有两个列表:文件 1 有 200000 行,看起来像
MAP2K4 FLNC
MYPN ACTN2
ACVR1 FNTA
UGT2A1 HPGDS
RPA2 STAT3
ARF1 GGA3
ARF3 ARFIP2
ARF3 ARFIP1
AKR1A1 EXOSC4
RPA2 GAS7
APP APPBP2
APLP1 DAB1
CITED2 TFAP2A
EP300 TFAP2A
APOB MTTP
ARRB2 RALGDS
ARRB2 ZNF807
文件 2 有 700000 行,如下所示:
MAP2K4 FLNC
MAP2K4 rs10036867
MAP2K4 ACTN2
MAP2K4 TEP1
ACTN2 MYPN
UGT2A1 NDUFAF6
RPA2 rs10109257
RPA2 rs10151961
GAS7 RPA2
APOB PDZRN4
APOB BICD1
ARRB2 ZNF807
ARRB2 FAM107B
尽管元素的顺序不同,我仍需要获取这两个列表之间的匹配行。例如,在上面的示例中,它应该如下所示:
MAP2K4 FLNC
ACTN2 MYPN
RPA2 GAS7
ARRB2 ZNF807
我写了以下内容,但这似乎需要永远!
col0_file1 = []
col1_file1 = []
col0_file2 = []
col1_file2 = []
with open('File1') as f1, open('File2') as f2:
for line in f1:
col0,col1 = line.split()
col0_file1.append(col0)
col1_file1.append(col1)
for line in f2:
col0,col1 = line.split()
col0_file2.append(col0)
col1_file2.append(col1)
result = []
for x in range(len(col0_file1)):
for i, j in map(None, col0_file2, col1_file2):
if i == col0_file1[x] and j == col1_file1[x]:
result.append([i,j])
elif j == col0_file1[x] and i == col1_file[x]:
result.append([i,j])
with open('matching', 'w') as out:
for elem in result:
out.write('{a} \n'.format(a = '\t'.join(elem)))
有什么办法可以简化复杂性?还是更好的方法?