Question

让字符串为"AAAGQWERTYUIOPAGCTHJKLAAAGZXCVBNMAGCT"。我想找到AAAG和AGCT之间的字符串。

我希望输出为["QWERTYUIOP","ZXCVBNM"]，即字符串列表。

如何使用正则表达式或类似技术来做到这一点？

我尝试过

def find_distances_between_motifs(positions1, positions2, motif_length1):
diff1 = []
diff2 = []
pos2 = 0
flag = 0
for pos1 in range(len(positions1)):
    if pos2 >= len(positions2):
        break
    if flag == 1:
        flag = 0
        pos1 -= 1
    if positions2[pos2] - positions1[pos1] > 30:
        diff1.append(NaN)
        diff2.append(NaN)
        continue
    elif positions2[pos2] - positions1[pos1] < 1:
        pos2 += 1
        diff2.append(NaN)
        flag = 1
    elif pos1 < len(positions1) - 1 and positions1[pos1+1] > positions2[pos2]:
        diff1.append(positions[pos2] - positions[pos1] - motif_length1)
        diff2.append(pos2)
        pos2 += 1
    else:
        continue
return diff1, diff2

我想返回两个数组-一个数组的位置在两个基序之间，而第二个数组的第二个基序将给出先前的距离。

Answer 1

使用正则表达式。 re.findall和Lookbehind＆Lookahead

例如：

import re
s = "AAAGQWERTYUIOPAGCTHJKLAAAGZXCVBNMAGCT"
print( re.findall(r"(?<=AAAG).*?(?=AGCT)", s))

输出：

['QWERTYUIOP', 'ZXCVBNM']

Answer 2

如果您不想使用正则表达式，那么我已经编写了代码。它有点复杂，但是如果您仔细研究它，那么您将会理解。

def addd(llist,word,word2):
    xx1 = sum([[i, word] for i in llist], [])[:-1]
    try:iii = xx1.index(word2);del xx1[iii]
    except:pass
    return xx1
a,output = addd("AAAGQWERTYUIOPAGCTHJKLAAAGZXCVBNMAGCT".split("AAAG"),"St4rT",""),[]
for i,x in enumerate(a):
    if "AGCT" in x:
        output.append(addd(x.split("AGCT"),"3nD.",""))
    else:output.append(x)
total = []
for i in output:
    if isinstance(i,list):total+=i
    elif isinstance(i,str):total.append(i)
output,typ = [],0
for x,i in enumerate(total):
    if typ == 0 and i == "St4rT":
        try:output.append(total[x+1]);typ = 1
        except:pass
    elif typ == 1 and i == "3nD.":typ = 0
print(output)

输出：

['QWERTYUIOP', 'ZXCVBNM']

如何在python字符串中的子字符串之间找到子字符串？

2 个答案: