1

所以我有下面的脚本：

def single_to_tripple(res):
    aa = {'R':'ARG','H':'HIS','K':'LYS','D':'ASP','E':'GLU','S':'SER','T':'THR','N':'ASN','Q':'GLN','C':'CYS','U':'SEC','G':'GLY','P':'PRO','A':'ALA','I':'ILE','L':'LEU','M':'MET','F':'PHE','W':'TRP','Y':'TYR','V':'VAL'}
    return(aa[res])
seq = 'ASALKDYYAIMGVKPTDDLKTIKTAYRRLARKYHPDVSKEPDAEARFKEVAEAWEVLSDEQRRAEYDQMWQHRNDPQFNRQFHHGDGQSFNAEDFDDIFSSIFGQHARQSRQRPATRGHDIEIEVAVFLEETLTEHKRTISYNLPVYNAFGMIEQEIPKTLNVKIPAGVGNGQRIRLKGQGTPGENGGPNGDLWLVIHIAPHPLFDIVGQDLEIVVPVSPWEAALGAKVTVPTLKESILLTIPPGSQAGQRLRVKGKGLVSKKQTGDLYAVLKIVMPPKPDENTAALWQQLADAQSSFDPRKDWGKA'
length = len(seq)

for i,v in enumerate(xrange(0,len(seq),13)):
    line = seq[v:v+13]
    out_line = ('{:<3} '*13).format(single_to_tripple(line[0]),single_to_tripple(line[1]),single_to_tripple(line[2]),single_to_tripple(line[3]),single_to_tripple(line[4]),single_to_tripple(line[5]),single_to_tripple(line[6]),single_to_tripple(line[7]),single_to_tripple(line[8]),single_to_tripple(line[9]),single_to_tripple(line[10]),single_to_tripple(line[11]),single_to_tripple(line[12]))
    print out_line

我正在使用脚本seq每 13 个元素拼接一次字符串，然后将拼接字符串中的每个元素从其单字母代码转换为single_to_tripple. 我的数据输出需要包含由空格分隔的 13 列。如果接头不包含 13 个元素，则会在最后一个接头处出现问题。我怎样才能捕捉到这个并像往常一样格式化字符串？

我enumerate在 for 循环中使用，因为稍后我需要添加行号。

我当前的代码输出：

ALA SER ALA LEU LYS ASP TYR TYR ALA ILE MET GLY VAL 
LYS PRO THR ASP ASP LEU LYS THR ILE LYS THR ALA TYR 
ARG ARG LEU ALA ARG LYS TYR HIS PRO ASP VAL SER LYS 
GLU PRO ASP ALA GLU ALA ARG PHE LYS GLU VAL ALA GLU 
ALA TRP GLU VAL LEU SER ASP GLU GLN ARG ARG ALA GLU 
TYR ASP GLN MET TRP GLN HIS ARG ASN ASP PRO GLN PHE 
ASN ARG GLN PHE HIS HIS GLY ASP GLY GLN SER PHE ASN 
ALA GLU ASP PHE ASP ASP ILE PHE SER SER ILE PHE GLY 
GLN HIS ALA ARG GLN SER ARG GLN ARG PRO ALA THR ARG 
GLY HIS ASP ILE GLU ILE GLU VAL ALA VAL PHE LEU GLU 
GLU THR LEU THR GLU HIS LYS ARG THR ILE SER TYR ASN 
LEU PRO VAL TYR ASN ALA PHE GLY MET ILE GLU GLN GLU 
ILE PRO LYS THR LEU ASN VAL LYS ILE PRO ALA GLY VAL 
GLY ASN GLY GLN ARG ILE ARG LEU LYS GLY GLN GLY THR 
PRO GLY GLU ASN GLY GLY PRO ASN GLY ASP LEU TRP LEU 
VAL ILE HIS ILE ALA PRO HIS PRO LEU PHE ASP ILE VAL 
GLY GLN ASP LEU GLU ILE VAL VAL PRO VAL SER PRO TRP 
GLU ALA ALA LEU GLY ALA LYS VAL THR VAL PRO THR LEU 
LYS GLU SER ILE LEU LEU THR ILE PRO PRO GLY SER GLN 
ALA GLY GLN ARG LEU ARG VAL LYS GLY LYS GLY LEU VAL 
SER LYS LYS GLN THR GLY ASP LEU TYR ALA VAL LEU LYS 
ILE VAL MET PRO PRO LYS PRO ASP GLU ASN THR ALA ALA 
LEU TRP GLN GLN LEU ALA ASP ALA GLN SER SER PHE ASP 
Traceback (most recent call last):
  File "make_seq_res.py", line 10, in <module>
    out_line = ('{:<3} '*13).format(single_to_tripple(line[0]),single_to_tripple(line[1]),single_to_tripple(line[2]),single_to_tripple(line[3]),single_to_tripple(line[4]),single_to_tripple(line[5]),single_to_tripple(line[6]),single_to_tripple(line[7]),single_to_tripple(line[8]),single_to_tripple(line[9]),single_to_tripple(line[10]),single_to_tripple(line[11]),single_to_tripple(line[12]))
IndexError: string index out of range
4

3 回答 3

3

您必须手动输入这么多变量的事实应该给您一个提示，即您正在做的事情超出了产生该输出的必要性。

在不更改大部分原始代码的情况下，可以这样做：

for i,v in enumerate(xrange(0,len(seq),13)):
    line = seq[v:v+13]
    out_line = ' '.join('{:<3}'.format(single_to_tripple(part)) for part in line)
    print out_line

正如 Martijn 指出的那样，三元组始终是三个字符，因此您实际上可以跳过格式：

out_line = ' '.join(single_to_tripple(part) for part in line)
于 2013-03-06T12:06:18.390 回答
2

您只需要将字符串连接在一起，无需格式化：

for i,v in enumerate(xrange(0,len(seq),13)):
    line = seq[v:v+13]
    print ' '.join([single_to_tripple(part) for part in line])

这里没有必要使事情过于复杂。:-)

请注意，使用 时str.join()，请在生成器表达式上使用列表推导（因此包括[...]），因为无论如何.join()都会强制转换为列表，从而使列表推导更快。

结果（最后 3 行）：

ILE VAL MET PRO PRO LYS PRO ASP GLU ASN THR ALA ALA
LEU TRP GLN GLN LEU ALA ASP ALA GLN SER SER PHE ASP
PRO ARG LYS ASP TRP GLY LYS ALA

您还可以使用itertools-based grouper 来简化循环：

from itertools import izip_longest

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return izip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

aa = {'R':'ARG','H':'HIS','K':'LYS','D':'ASP','E':'GLU','S':'SER','T':'THR','N':'ASN','Q':'GLN','C':'CYS','U':'SEC','G':'GLY','P':'PRO','A':'ALA','I':'ILE','L':'LEU','M':'MET','F':'PHE','W':'TRP','Y':'TYR','V':'VAL', None: ''}
def single_to_tripple(res):
    return(aa[res])

for line in grouper(13, seq):
    print ' '.join([single_to_tripple(part) for part in line])

我single_to_tripple()通过将映射移出函数（无需每次调用​​它时都定义它）并添加一个键None（grouper 用None值填充最后一组）来增强您的函数。

于 2013-03-06T12:09:24.473 回答
0

您可以保存您的行长，而不是使用它：

def single_to_tripple(res):
    aa = {'R':'ARG','H':'HIS','K':'LYS','D':'ASP','E':'GLU','S':'SER','T':'THR','N':'ASN','Q':'GLN','C':'CYS','U':'SEC','G':'GLY','P':'PRO','A':'ALA','I':'ILE','L':'LEU','M':'MET','F':'PHE','W':'TRP','Y':'TYR','V':'VAL'}
    return(aa[res])

seq = 'ASALKDYYAIMGVKPTDDLKTIKTAYRRLARKYHPDVSKEPDAEARFKEVAEAWEVLSDEQRRAEYDQMWQHRNDPQFNRQFHHGDGQSFNAEDFDDIFSSIFGQHARQSRQRPATRGHDIEIEVAVFLEETLTEHKRTISYNLPVYNAFGMIEQEIPKTLNVKIPAGVGNGQRIRLKGQGTPGENGGPNGDLWLVIHIAPHPLFDIVGQDLEIVVPVSPWEAALGAKVTVPTLKESILLTIPPGSQAGQRLRVKGKGLVSKKQTGDLYAVLKIVMPPKPDENTAALWQQLADAQSSFDPRKDWGKA'
length = len(seq)

for i,v in enumerate(xrange(0,len(seq),13)):
    line = seq[v:v+13]
    length = len(line)
    out_line = ('{:<3} '*length).format(*[single_to_tripple(a) for a in line])
    print out_line
于 2013-03-06T12:08:23.380 回答