使用 python 进行复杂字符串过滤
Complex string filtering with python
我有一个很长的字符串，它是一个系统发育树，我想做一个非常具体的过滤。
(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;
基本上每一个x@y就是一个species@gene_id的信息。我想做的是减少它，这样我就只有 x 而不是 x@y。
(Esy, Aar,(Spa,Cpl))...
我尝试先拆分字符串，但问题是字符串对于我想要实现的目标有不同的 'split points'，即某些部分 x@y 以 , 结尾，而其他部分以 , 结尾)。我搜索了一个解决方案并看到了正则表达式操作，但我是 Python 的新手，我不确定这是否是我应该关注的。我也考虑过 strip() 但似乎我需要为此指定要删除的字符。
主要问题是没有'pattern'让我告诉Python跟随。唯一的问题是所有物种 ID 都是 3 个字母，并且它们在 @ 字符之前。
有什么方法可以达到我的要求吗？如果你能帮助我解决我的问题，我将非常高兴。提前致谢。
这种功能怎么样：
def parse_string(string):
new_string = ''
skip = False
for char in string:
if char == '@':
skip = True
if char == ',':
skip = False
if not skip or char in ['(', ')']:
new_string += char
return new_string
在您的字符串上调用它：
string = '(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;'
parse_string(string)
> '(Esy,Aar,((Spa,Cpl),(((Bst,((Aly,Ath),(Chi,Cru))),(((Hco,Hlo),Hla),Hse)),(Esa,Aal))))'
试一试：
import re:
pat = re.compile(r'(\w{3})@')
txt = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
pat.findall(t)
结果：
['Esy', 'Aar', 'Spa', 'Cpl', 'Bst', 'Aly', 'Ath', 'Chi', 'Cru', 'Hco', 'Hlo', 'Hla', 'Hse', 'Esa', 'Aal']
如果您需要完整的结构，我们可以尝试删除不需要的部分：
pat = re.compile(r'(@|:)[^/),]*')
pat.sub('',t).replace(',', ', ')
结果：
'(Esy, Aar, ((Spa, Cpl), (((Bst, ((Aly, Ath), (Chi, Cru))), (((Hco, Hlo), Hla), Hse)), (Esa, Aal))))'
因为您正在尝试解析系统发育树，所以我强烈建议让 BioPython 为您完成繁重的工作。
您可以使用 Bio.Phylo 轻松解析和显示系统发育。然后它只是遍历所有树元素并在 'at' 符号处拆分名称。
因为 Phylo 期望输入在文件中，所以我们使用 io.StringIO 创建一个内存中的类文件对象。获得完整的树就像
一样简单
Phylo.read(io.StringIO(s), 'newick')
为了检查解析后的树是否正常，我用 print(tree) 打印了一次。
现在我们要更改所有包含 '@' 的节点名称。使用 tree.find_elements 我们可以访问所有节点。有些节点没有名称，有些节点可能不包含 '@'。所以要格外小心，我们首先检查 if n.name and '@' in n.name。只有这样，我们才能在 '@' 处拆分每个节点的名称，并只取它的第一部分（索引 0）：
n.name = n.name.split('@')[0]
为了重新创建初始字符串表示，我们使用 Phylo.write:
out = io.StringIO()
Phylo.write(tree, out, "newick")
print(out.getvalue())
同样，write 想要获取文件参数 - 如果我们只想获取字符串，我们可以再次使用 StringIO 对象。
完整代码：
import io
from Bio import Phylo
if __name__ == '__main__':
s = '(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;'
tree = Phylo.read(io.StringIO(s), 'newick')
print(' before '.center(20, '='))
print(tree)
for n in tree.find_elements():
if n.name and '@' in n.name:
n.name = n.name.split('@')[0]
print(' result '.center(20, '='))
out = io.StringIO()
Phylo.write(tree, out, "newick")
print(out.getvalue())
输出：
====== before ======
Tree(rooted=False, weight=1.0)
Clade(branch_length=0.0129090235079)
Clade(branch_length=0.0726396855636, name='Esy@ESY15_g64743_DN3_SP7_c0')
Clade(branch_length=0.137507902808, name='Aar@AA_maker7399_1')
Clade(branch_length=0.0129090235079)
Clade(branch_length=9.05326020871e-05)
Clade(branch_length=0.0318934795022, name='Spa@Tp2g18720')
Clade(branch_length=0.0273465005242, name='Cpl@CP2_g48793_DN3_SP8_c')
Clade(branch_length=0.00328120860999)
Clade(branch_length=0.00859075940423)
Clade(branch_length=0.0340484449097)
Clade(branch_length=0.0332592496158, name='Bst@Bostr_13083s0053_1')
Clade(branch_length=0.0150356382287)
Clade(branch_length=0.0205924636564)
Clade(branch_length=0.0328569260951, name='Aly@AL8G21130_t1')
Clade(branch_length=0.0391706378372, name='Ath@AT5G48370_1')
Clade(branch_length=0.00998579652059)
Clade(branch_length=0.0954469923893, name='Chi@CARHR183840_1')
Clade(branch_length=0.0570981548016, name='Cru@Carubv10026342m')
Clade(branch_length=0.0372829371381)
Clade(branch_length=0.0206478928557)
Clade(branch_length=0.0144626717872)
Clade(branch_length=0.00823215335663, name='Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100')
Clade(branch_length=0.0085462978729, name='Hlo@DN13684_c0_g1_i1_p1')
Clade(branch_length=0.0225079453622, name='Hla@DN22821_c0_g1_i1_p1')
Clade(branch_length=0.048590776459, name='Hse@DN23412_c0_g1_i3_p1')
Clade(branch_length=1.00000050003e-06)
Clade(branch_length=0.0378509854703, name='Esa@Thhalv10004228m')
Clade(branch_length=0.0712272454125, name='Aal@Aa_G102140_t1')
==== result =====
(Esy:0.07264,Aar:0.13751,((Spa:0.03189,Cpl:0.02735):0.00009,(((Bst:0.03326,((Aly:0.03286,Ath:0.03917):0.02059,(Chi:0.09545,Cru:0.05710):0.00999):0.01504):0.03405,(((Hco:0.00823,Hlo:0.00855):0.01446,Hla:0.02251):0.02065,Hse:0.04859):0.03728):0.00859,(Esa:0.03785,Aal:0.07123):0.00000):0.00328):0.01291):0.01291;
Phylo 的默认格式使用的数字少于原始树中的数字。为了保持数字不变，只需用“%s”覆盖分支长度格式字符串：
Phylo.write(tree, out, "newick", format_branch_length="%s")
如果您需要输出中的方括号，请尝试使用此正则表达式：
import re
regex = r"@[A-Za-z0-9_\.:]+|[0-9:\.;e-]+"
phylogenetic_tree = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
print(re.sub(regex,"",phylogenetic_tree))
输出：
(Esy,Aar,((Spa,Cpl),(((Bst,((Aly,Ath),(Chi,Cru))),(((Hco,Hlo),Hla),Hs)),(Esa,Aal))))
你可以使用正则表达式：
import re
s = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
p = "...?(?=@)|\(|\)"
result = re.findall(p, s)
你的结果是一个列表，所以你可以把它变成字符串或用它做任何事情
解释发生了什么：
p 是正则表达式模式
所以在这个模式中：
.表示匹配任意单词
...?(?=@) 意味着匹配任何单词，直到我得到一个单词 ? wich ? 是 @，所以整个模式意味着你在 @ 之前得到任何三个单词
| 是 or 语句，我在这里用它来寻找另一个模式
剩下的就是找)和(
解析代码可能很难理解。 Tatsu 让您通过组合语法和 python:
编写可读的解析代码
text = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
import sys
import tatsu
grammar = """
start = things ';'
;
things = thing [ ',' things ]
;
thing = x '@' y ':' number
| '(' things ')' ':' number
;
x = /\w+/
;
y = /\w+/
;
number = /[+-]?\d+\.?\d*(e?[+-]?\d*)/
;
"""
class Semantics:
def x(self, ast):
# the method name matches the rule name
print('X =', ast)
parser = tatsu.compile(grammar, semantics=Semantics())
parser.parse(text)
我有一个很长的字符串，它是一个系统发育树，我想做一个非常具体的过滤。
(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;
基本上每一个x@y就是一个species@gene_id的信息。我想做的是减少它，这样我就只有 x 而不是 x@y。
(Esy, Aar,(Spa,Cpl))...
我尝试先拆分字符串，但问题是字符串对于我想要实现的目标有不同的 'split points'，即某些部分 x@y 以 , 结尾，而其他部分以 , 结尾)。我搜索了一个解决方案并看到了正则表达式操作，但我是 Python 的新手，我不确定这是否是我应该关注的。我也考虑过 strip() 但似乎我需要为此指定要删除的字符。
主要问题是没有'pattern'让我告诉Python跟随。唯一的问题是所有物种 ID 都是 3 个字母，并且它们在 @ 字符之前。
有什么方法可以达到我的要求吗？如果你能帮助我解决我的问题，我将非常高兴。提前致谢。
这种功能怎么样：
def parse_string(string):
new_string = ''
skip = False
for char in string:
if char == '@':
skip = True
if char == ',':
skip = False
if not skip or char in ['(', ')']:
new_string += char
return new_string
在您的字符串上调用它：
string = '(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;'
parse_string(string)
> '(Esy,Aar,((Spa,Cpl),(((Bst,((Aly,Ath),(Chi,Cru))),(((Hco,Hlo),Hla),Hse)),(Esa,Aal))))'
试一试：
import re:
pat = re.compile(r'(\w{3})@')
txt = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
pat.findall(t)
结果：
['Esy', 'Aar', 'Spa', 'Cpl', 'Bst', 'Aly', 'Ath', 'Chi', 'Cru', 'Hco', 'Hlo', 'Hla', 'Hse', 'Esa', 'Aal']
如果您需要完整的结构，我们可以尝试删除不需要的部分：
pat = re.compile(r'(@|:)[^/),]*')
pat.sub('',t).replace(',', ', ')
结果：
'(Esy, Aar, ((Spa, Cpl), (((Bst, ((Aly, Ath), (Chi, Cru))), (((Hco, Hlo), Hla), Hse)), (Esa, Aal))))'
因为您正在尝试解析系统发育树，所以我强烈建议让 BioPython 为您完成繁重的工作。
您可以使用 Bio.Phylo 轻松解析和显示系统发育。然后它只是遍历所有树元素并在 'at' 符号处拆分名称。
因为 Phylo 期望输入在文件中，所以我们使用 io.StringIO 创建一个内存中的类文件对象。获得完整的树就像
Phylo.read(io.StringIO(s), 'newick')
为了检查解析后的树是否正常，我用 print(tree) 打印了一次。
现在我们要更改所有包含 '@' 的节点名称。使用 tree.find_elements 我们可以访问所有节点。有些节点没有名称，有些节点可能不包含 '@'。所以要格外小心，我们首先检查 if n.name and '@' in n.name。只有这样，我们才能在 '@' 处拆分每个节点的名称，并只取它的第一部分（索引 0）：
n.name = n.name.split('@')[0]
为了重新创建初始字符串表示，我们使用 Phylo.write:
out = io.StringIO()
Phylo.write(tree, out, "newick")
print(out.getvalue())
同样，write 想要获取文件参数 - 如果我们只想获取字符串，我们可以再次使用 StringIO 对象。
完整代码：
import io
from Bio import Phylo
if __name__ == '__main__':
s = '(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;'
tree = Phylo.read(io.StringIO(s), 'newick')
print(' before '.center(20, '='))
print(tree)
for n in tree.find_elements():
if n.name and '@' in n.name:
n.name = n.name.split('@')[0]
print(' result '.center(20, '='))
out = io.StringIO()
Phylo.write(tree, out, "newick")
print(out.getvalue())
输出：
====== before ======
Tree(rooted=False, weight=1.0)
Clade(branch_length=0.0129090235079)
Clade(branch_length=0.0726396855636, name='Esy@ESY15_g64743_DN3_SP7_c0')
Clade(branch_length=0.137507902808, name='Aar@AA_maker7399_1')
Clade(branch_length=0.0129090235079)
Clade(branch_length=9.05326020871e-05)
Clade(branch_length=0.0318934795022, name='Spa@Tp2g18720')
Clade(branch_length=0.0273465005242, name='Cpl@CP2_g48793_DN3_SP8_c')
Clade(branch_length=0.00328120860999)
Clade(branch_length=0.00859075940423)
Clade(branch_length=0.0340484449097)
Clade(branch_length=0.0332592496158, name='Bst@Bostr_13083s0053_1')
Clade(branch_length=0.0150356382287)
Clade(branch_length=0.0205924636564)
Clade(branch_length=0.0328569260951, name='Aly@AL8G21130_t1')
Clade(branch_length=0.0391706378372, name='Ath@AT5G48370_1')
Clade(branch_length=0.00998579652059)
Clade(branch_length=0.0954469923893, name='Chi@CARHR183840_1')
Clade(branch_length=0.0570981548016, name='Cru@Carubv10026342m')
Clade(branch_length=0.0372829371381)
Clade(branch_length=0.0206478928557)
Clade(branch_length=0.0144626717872)
Clade(branch_length=0.00823215335663, name='Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100')
Clade(branch_length=0.0085462978729, name='Hlo@DN13684_c0_g1_i1_p1')
Clade(branch_length=0.0225079453622, name='Hla@DN22821_c0_g1_i1_p1')
Clade(branch_length=0.048590776459, name='Hse@DN23412_c0_g1_i3_p1')
Clade(branch_length=1.00000050003e-06)
Clade(branch_length=0.0378509854703, name='Esa@Thhalv10004228m')
Clade(branch_length=0.0712272454125, name='Aal@Aa_G102140_t1')
==== result =====
(Esy:0.07264,Aar:0.13751,((Spa:0.03189,Cpl:0.02735):0.00009,(((Bst:0.03326,((Aly:0.03286,Ath:0.03917):0.02059,(Chi:0.09545,Cru:0.05710):0.00999):0.01504):0.03405,(((Hco:0.00823,Hlo:0.00855):0.01446,Hla:0.02251):0.02065,Hse:0.04859):0.03728):0.00859,(Esa:0.03785,Aal:0.07123):0.00000):0.00328):0.01291):0.01291;
Phylo 的默认格式使用的数字少于原始树中的数字。为了保持数字不变，只需用“%s”覆盖分支长度格式字符串：
Phylo.write(tree, out, "newick", format_branch_length="%s")
如果您需要输出中的方括号，请尝试使用此正则表达式：
import re
regex = r"@[A-Za-z0-9_\.:]+|[0-9:\.;e-]+"
phylogenetic_tree = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
print(re.sub(regex,"",phylogenetic_tree))
输出：
(Esy,Aar,((Spa,Cpl),(((Bst,((Aly,Ath),(Chi,Cru))),(((Hco,Hlo),Hla),Hs)),(Esa,Aal))))
你可以使用正则表达式：
import re
s = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
p = "...?(?=@)|\(|\)"
result = re.findall(p, s)
你的结果是一个列表，所以你可以把它变成字符串或用它做任何事情
解释发生了什么：
p 是正则表达式模式
所以在这个模式中：
.表示匹配任意单词
...?(?=@) 意味着匹配任何单词，直到我得到一个单词 ? wich ? 是 @，所以整个模式意味着你在 @ 之前得到任何三个单词
| 是 or 语句，我在这里用它来寻找另一个模式
剩下的就是找)和(
解析代码可能很难理解。 Tatsu 让您通过组合语法和 python:
编写可读的解析代码text = "(Esy@ESY15_g64743_DN3_SP7_c0:0.0726396855636,Aar@AA_maker7399_1:0.137507902808,((Spa@Tp2g18720:0.0318934795022,Cpl@CP2_g48793_DN3_SP8_c:0.0273465005242):9.05326020871e-05,(((Bst@Bostr_13083s0053_1:0.0332592496158,((Aly@AL8G21130_t1:0.0328569260951,Ath@AT5G48370_1:0.0391706378372):0.0205924636564,(Chi@CARHR183840_1:0.0954469923893,Cru@Carubv10026342m:0.0570981548016):0.00998579652059):0.0150356382287):0.0340484449097,(((Hco@scaff1034_g23864_DN3_SP8_c_TE35_CDS100:0.00823215335663,Hlo@DN13684_c0_g1_i1_p1:0.0085462978729):0.0144626717872,Hla@DN22821_c0_g1_i1_p1:0.0225079453622):0.0206478928557,Hse@DN23412_c0_g1_i3_p1:0.048590776459):0.0372829371381):0.00859075940423,(Esa@Thhalv10004228m:0.0378509854703,Aal@Aa_G102140_t1:0.0712272454125):1.00000050003e-06):0.00328120860999):0.0129090235079):0.0129090235079;"
import sys
import tatsu
grammar = """
start = things ';'
;
things = thing [ ',' things ]
;
thing = x '@' y ':' number
| '(' things ')' ':' number
;
x = /\w+/
;
y = /\w+/
;
number = /[+-]?\d+\.?\d*(e?[+-]?\d*)/
;
"""
class Semantics:
def x(self, ast):
# the method name matches the rule name
print('X =', ast)
parser = tatsu.compile(grammar, semantics=Semantics())
parser.parse(text)