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如何在python字符串中的子字符串之间找到子字符串?

bioinformatics string python

Skywalker · 技术社区 · 6 年前

让这根绳子 "AAAGQWERTYUIOPAGCTHJKLAAAGZXCVBNMAGCT" . 我想找到AAAG和AGCT之间的字符串。

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

我试过了

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-一个是序列长度在基序之间的位置,第二个是第二个基序的位置,这将给出前面的距离。

2 回复 | 直到 6 年前

Rakesh 6 年前

使用正则表达式。 re.findall 带lookback和Lookahead

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

['QWERTYUIOP', 'ZXCVBNM']

Nouman ProfHase85 6 年前

如果你不想使用正则表达式,那么我已经做了一个代码。这有点复杂,但如果你仔细看看,然后你就会明白。

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)