我有两个小序列,我在“长字符串”中搜索。如果两个序列都找到,则“长字符串”的键将附加到列表中(我搜索的字符串是字典值)。

现在我正在寻找一种方法,获取/计算两个子串之间的距离(如果找到了)。

例如:

String: ABCDEFGHIJKL
sequence1: ABC
sequence2: JKL

我想得到DEFGHI的长度,是6。

这是我查找子字符串的代码,其中包含一些我想要的“伪codish”概念(变量开始和结束)。此代码无效(ofc)

def search (myDict, list1, list2):
    # initialize empty list to store found keys
    a=[]
    # iterating through dictionary
    for key, value in myDict.items():
        # if -35nt motif is found between -40 and -20
        for item in thirtyFive:
            if item in value[60:80]:
                start=myDict[:item]
            # it is checked for the -10nt motif from -40 to end
                for item in ten:
                    if item in value[80:]:
                        end=myDict[:item]
                # if both conditions are true, the IDs are
                # appended to the list
                        a.append(key)
    distance=start-end
    return a, distance

第二个想法:

所以,我想知道,如果我的第一个想法,在我找到小序列的时候,以某种方式做到这一点,是否有可能,如果我用我的第二个想法朝着正确的方向思考。

提前感谢:)

使用str.find方法在@Carlos之后求解

def search (myDict, list1, list2):
    # initialize empty list to store found keys
    a=[]
    # iterating through dictionary
    for key, value in myDict.items():
        # if -35nt motif is found between -40 and -20
        for item in thirtyFive:
            if item in value[60:80]:
                start=value.find(item)
            # it is checked for the -10nt motif from -20 to end
                for item in ten:
                    if item in value[80:]:
                        end=value.find(item)
                # if both conditions are true, the IDs are
                # appended to the list
                        a.append(key)
                        search.distance=end-start-len(item)

    return a

# calling search function
x=search(d,thirtyFive,ten)
#some other things I need to print
y=len(x)
print(str(x))
print(y)
# desired output
print(search.distance)