我实际上有一个巨大的multifasta seq文件，例如：
>Seq_1_0035_0035
ATTGGAT
>Seq_2_0042_0035
ATTGAGGA
>EOGWX56TR_0035_0042 (busco)
ATGGAGAT
>EOGWX56TR_0042_0042 (busco)
ATGGATGG
>Seq6_035_0042
ATGGGAATAG
>EOG55FTG_0035_0042 (busco)
AATGGATA
>EOG5GFFTA_0042_0042 (busco)
ATGGAGATA
>Seq56_0035_0042
ATGGAGATAT
>EOGATTT_0035_0042 (busco)
AAATGAGATA
>EOGATTT_0042_0042 (busco)
ATGGAAT
>EOGATTA_0042_0042 (busco)
ATAGGAGAT
我实际上想计算文件中有多少Busco基因（它们都以
>EOG开头）。为此，我有一个脚本：count=1
for record in SeqIO.parse("concatenate_with_busco_names_0035_0042_aa.fa", "fasta"):
count+=1
print(count)
set_of_labels = set()
with open("concatenate_with_busco_names_0035_0042_aa.fa") as f:
for line in f:
if line.startswith('>EOG'):
label = line[4:].split('_')[0]
set_of_labels.add(label)
print("Total number of Busco genes: " + str(len(set_of_labels)))
但是我也想知道每个对应的组件之间我有多少个基因。我会更好地解释；
如您所见，每个seqID
such _number_number中都有两个数字这些数字是特定的，第一个
_number对应于该序列所属的物种，第二个_number对应其特定编号。无论如何，我想是否可以像我一样将第一个数字
_0035和_0042的seq获得多少个Busco基因并且
seq ID多少个：
_0035_0042
_0035_0042
_0042_0042
_0042_0035
在以上示例中，它将是：
Total busco: 5 (I count only once if the >busco is present even if _number are different)
Total busco for the specie _0035 (_0035_0042 and _0035_0035) : 3
Total busco for the specie _0042 (_0042_0042 and _0042_0035) : 4
Total busco for the specific specie _0035_0042 : 3
Total busco for the specific specie _0042_0035 : 0
Total busco for the specific specie _0042_0042 : 4
Total busco for the specific specie _0035_0035 : 0
您好希望很清楚，实际上第一部分（
total busco:）已经由我的脚本完成，我只需要数一下其他7种方式。这是真实数据data
最佳答案
除了busco计数器外，您还可以使用多个计数器来获取物种和特定物种的单独计数，例如：
import collections
busco = collections.defaultdict(int) # busco counter
species = collections.defaultdict(int) # species counter
specific_species = collections.defaultdict(int) # specific species counter
with open("concatenate_with_busco_names_0035_0042_aa.fa", "r") as f:
for line in f:
if line[:4] == ">EOG":
entry = line.split()[0][4:].split('_')
busco[entry[0]] += 1
species[entry[1]] += 1
specific_species[entry[1] + "_" + entry[2]] += 1
print("Total busco: {}".format(len(busco)))
for specie, total in species.items():
print("Total busco for the specie {}: {}".format(specie, total))
for specie, total in specific_species.items():
print("Total busco for the specific specie {}: {}".format(specie, total))
哪个应该产生：
Total busco: 5 Total busco for the specie 0035: 3 Total busco for the specie 0042: 4 Total busco for the specific specie 0035_0042: 3 Total busco for the specific specie 0042_0042: 4
Non-listed (specific) species won't show up but if you really want to print them out you can combine them from the species counter and print their values (default is 0):
import itertools
print("Total busco: {}".format(len(busco)))
for specie, total in species.items():
print("Total busco for the specie {}: {}".format(specie, total))
for specie in itertools.product(species, species):
s = "_".join(specie)
print("Total busco for the specific specie {}: {}".format(s, specific_species[s]))
产生：
Total busco: 5 Total busco for the specie 0035: 3 Total busco for the specie 0042: 4 Total busco for the specific specie 0035_0035: 0 Total busco for the specific specie 0035_0042: 3 Total busco for the specific specie 0042_0035: 0 Total busco for the specific specie 0042_0042: 4
UPDATE: If you're after unique counts per busco then you need to inverse the count to index on specie / specific specie and collect busco values in a set as their value. Then all you need is to get the length of each set, something like:
import collections
import itertools
busco = set()
species = collections.defaultdict(set)
specific_species = collections.defaultdict(set)
with open("concatenate_with_busco_names_0035_0042_aa.fa", "r") as f:
for line in f:
if line[:4] == ">EOG":
entry = line.split()[0][4:].split('_')
busco.add(entry[0])
species[entry[1]].add(entry[0])
specific_species[entry[1] + "_" + entry[2]].add(entry[0])
print("Total busco: {}".format(len(busco)))
for specie, buscos in species.items():
print("Total busco for the specie {}: {}".format(specie, len(buscos)))
for specie in itertools.product(species, species):
s = "_".join(specie)
print("Total busco for the specific specie {}: {}".format(s, len(specific_species[s])))
为您的完整数据打印：
巴士总数：421
现金0035的总busco：402
现金0042的总busco：397
特定品种的总busco 0035_0035：392
特定品种的总巴士总费用0035_0042：262
特定品种的总巴士总票数0042_0035：305
特定品种的总巴士总票数0042_0042：383
关于python - 计数seqID python中的特定模式，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/50349288/