Question

目前我的程序采用line1，例如＆＃34; taaaaaaaaaaNataggggggggggNccc＆＃34;并将剪切结束的1个字符，直到它与line2匹配，例如＆＃34; taaaaaaaaaaNcccggggggggggNccc＆＃34;一旦它们匹配，它就会将它们连接在一起形成第3行，但是如果它们不匹配则应该删除另一个字符。如何重复切割动作直到它们匹配并且已经制作了line3？我已经考虑过for和while循环，但我不确定如何陈述这个问题。关于这个程序的其他所有内容都可以正常工作，但是当它尝试匹配它们时如果它失败它就会停止并且不会再回去再次尝试修剪。

我已经尝试了下面的代码，其中magic（匹配）本质上是用于idnetfy 2行匹配多少的计数代码，如果低于8则应该重复切割。但是当它被使用时，它要求匹配和魔术在while循环之前说明，这在开始时是正确的，这会扰乱代码的其余部分。

while magic(matching) >=8:
    line3=line1+line2
        print ("Matching and merging has occured as shown below")
        print (line3)

感兴趣的代码如下：

       n = 0
        consec_matches = []
        chars = defaultdict(int)    
for k, group in groupby(zip(line1_u_i, line2_u_rev_comp_join_i), class_chars):
        elems = len(list(group))
        chars[k] += elems
        if k == 'match':
        consec_matches.append((n, n+elems-1))
    n += elems
print ("Print chars below")
print (chars)
print ("Print consec_matches below")
print (consec_matches)
print ([x for x in consec_matches if x[1]-x[0] >= 9])
print (" Matches longer than 10 below")
list = [x for x in consec_matches if x[1]-x[0] >= 9]
flatten_list= [x for y in list for x in y]
print (flatten_list)
print ("Flatterend list")
matching=[y[1] for y in list for x in y if x ==0 ]
print ("Matching list below")
print (matching)
magic = lambda matching: int(''.join(str(i) for i in matching) or 0)
print (" Print magic matching below")
print (magic(matching))
line2_u_rev_comp_join_i_l = line2_u_rev_comp_join_i[magic(matching):]
print ("Print line2_u_rev_comp_join_i_l type below")
print (type(line2_u_rev_comp_join_i_l))
print ("Print line2_u_rev_comp_join_i_l sequence below")
print (line2_u_rev_comp_join_i_l)
line2_u_rev_comp_join_i_l_str = ''.join(line2_u_rev_comp_join_i_l)
print ('List of line2 converted to string')
print ("List2 before as list below")
print (line2_u_rev_comp_join_i_l)
print ("Line 2 reprinted when string as below")
print (line2_u_rev_comp_join_i_l_str)
print (line1_u_i)
print ("Magic below")
print (magic)
if magic(matching) >=8:
    line3=line1_u_i+line2_u_rev_comp_join_i_l_str
    print ("Matching and merging has occured as shown below")
    print (line3)
else:
    continue

切割代码是：

line2_u_rev_comp_join_i = line2_u_rev_comp_join[1:]
line1_u_i = line1_u[:-1]

Answer 1

l1 = "taaaaaaaaaaNataggggggggggNccc"
l2 = "taaaaaaaaaaNcccggggggggggNccc"
l1_ten=l1[0:10] # first ten chars
l2_ten=l2[0:10] 
if l1_ten==l2_ten:
    print l1_ten+l2_ten
taaaaaaaaataaaaaaaaa

如果希望每个字符串中的字符在相同索引处相等。

l1 = "taaaaaaaaaaNataggggggggggNccc"
l2 = "taaaaaaaaaaNcccggggggggggNccc"
count = 0
slice = 0
new_s=''
while count < len(l1):
    if l1[slice]==l2[slice]: 
        new_s+= l1[slice]
    count += 1
    slice += 1
new_s
In [13]: new_s
Out[13]: 'taaaaaaaaaaNggggggggggNccc'

您可以使用for循环来实现相同的目标：

new_s1=""
for i in range(len(l1)):
    if l1[i] ==l2[i]:
        new_s+=l1[i]

我假设您使用的是相同长度的字符串

使程序继续操作直到创建行

1 个答案: