如何在一个字典中组合函数的单独字典输出?

问题描述 投票:1回答:2

对于我们的python项目,我们必须解决多个问题。然而,我们坚持这一点:

“编写一个函数,给定一个FASTA文件名,返回一个字典,其序列ID为键,元组为值。该值表示序列的最小和最大分子量(序列可能不明确)。”

import collections
    from Bio import Seq
    from itertools import product
    def ListMW(file_name):
        seq_records = SeqIO.parse(file_name, 'fasta',alphabet=generic_dna)
        for record in seq_records:
            dictionary = Seq.IUPAC.IUPACData.ambiguous_dna_values
            result = []
            for i in product(*[dictionary[j] for j in record]):
                result.append("".join(i))
                molw = []
            for sequence in result:
                molw.append(SeqUtils.molecular_weight(sequence))
            tuple= (min(molw),max(molw))
            if min(molw)==max(molw):
                dict={record.id:molw}
            else:
                dict={record.id:(min(molw), max(molw))}

            print(dict) 

使用此代码,我们设法获得此输出:

{'seq_7009': (6236.9764, 6367.049999999999)}
{'seq_418': (3716.3642000000004, 3796.4124000000006)}
{'seq_9143_unamb': [4631.958999999999]}
{'seq_2888': (5219.3359, 5365.4089)}
{'seq_1101': (4287.7417, 4422.8254)}
{'seq_107': (5825.695099999999, 5972.8073)}
{'seq_6946': (5179.3118, 5364.420900000001)}
{'seq_6162': (5531.503199999999, 5645.577399999999)}
{'seq_504': (4556.920899999999, 4631.959)}
{'seq_3535': (3396.1715999999997, 3446.1969999999997)}
{'seq_4077': (4551.9108, 4754.0073)}
{'seq_1626_unamb': [3724.3894999999998]}

正如您所看到的,这不是一本字典,而是彼此之下的多个字典。那么无论如何我们可以更改我们的代码或输入一个额外的命令来获得这种格式:

{'seq_7009': (6236.9764, 6367.049999999999),
'seq_418': (3716.3642000000004, 3796.4124000000006),
'seq_9143_unamb': (4631.958999999999),
'seq_2888': (5219.3359, 5365.4089),
'seq_1101': (4287.7417, 4422.8254),
'seq_107': (5825.695099999999, 5972.8073),
'seq_6946': (5179.3118, 5364.420900000001),
'seq_6162': (5531.503199999999, 5645.577399999999),
'seq_504': (4556.920899999999, 4631.959),
'seq_3535': (3396.1715999999997, 3446.1969999999997),
'seq_4077': (4551.9108, 4754.0073),
'seq_1626_unamb': (3724.3894999999998)}

或者在某种程度上设法明确它应该使用seq_ID ans键和分子量作为一个字典的值?

python dictionary
2个回答
2
投票

在for循环之前设置字典,然后在循环期间更新它,例如:

import collections
    from Bio import Seq
    from itertools import product
    def ListMW(file_name):
        seq_records = SeqIO.parse(file_name, 'fasta',alphabet=generic_dna)
        retDict = {}
        for record in seq_records:
            dictionary = Seq.IUPAC.IUPACData.ambiguous_dna_values
            result = []
            for i in product(*[dictionary[j] for j in record]):
                result.append("".join(i))
                molw = []
            for sequence in result:
                molw.append(SeqUtils.molecular_weight(sequence))
            tuple= (min(molw),max(molw))
            if min(molw)==max(molw):
                retDict[record.id] = molw
            else:
                retDict[record.id] = (min(molw), max(molw))}
            # instead of printing now, print in the end of your function / script
            # print(dict) 

现在,你在循环的每个转弯处设置一个新的字典,并打印出来。打印大量的dict只是代码的正常行为。


1
投票

你在每次迭代时创建一个带有1个条目的字典。

你想要:

  • 在循环之前定义一个dict变量(更好地使用dct以避免重用内置类型名称)
  • 在循环中重写dict的赋值

所以在循环之前:

dct = {}

并且在循环中(而不是你的if + dict =代码),在三元表达式中,min&max只计算一次:

minval = min(molw)
maxval = max(molw)
dct[record.id] = molw if minval == maxval else (minval,maxval)
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