Question

我一直在研究一个程序，我在编程时遇到了问题。

这是我的代码：

dict=([{'geneA': [10, 20]}, {'geneB': [12, 45]}, {'geneC': [36, 50]}],
      [{'geneD': [45, 90]}, {'geneT': [100, 200]}],
      [{'geneF': [15, 25]}, {'geneX': [67, 200]}, {'GeneZ': [234, 384]}])

所以我基本上设置的dict等于染色体1,2和3的数据。有没有什么方法可以在dict中显示这三个字符串的名称而不将其作为dict索引的一部分？

Answer 1

这里的主要问题是您没有声明dictionary而是tuple。

尝试：

>> dict= {

    'chromosome1' : [{'geneA': [10, 20]}, {'geneB': [12, 45]}, {'geneC': [36, 50]}],
    'chromosome2' : [{'geneD': [45, 90]}, {'geneT': [100, 200]}],
    'chromosome3' : [{'geneF': [15, 25]}, {'geneX': [67, 200]}, {'GeneZ': [234, 384]}]
 }


>> print(dict['chromosome1'])
[{'geneA': [10, 20]}, {'geneB': [12, 45]}, {'geneC': [36, 50]}]
>> print(dict['chromosome2'])
[{'geneD': [45, 90]}, {'geneT': [100, 200]}]
>> print(dict['chromosome3'])
[{'geneF': [15, 25]}, {'geneX': [67, 200]}, {'GeneZ': [234, 384]}]

POO方法

您也可以尝试POO方法。如果你实现了几个类，如下所示：

class Gene:
    def __init__(self, name, data):
        self.name = name
        self.data = data
    def __getitem__(self, index):
        return self.data[index]

class Chromosome:
    def __init__(self, name, data):
        self.name = name
        self.data = data
    def __getitem__(self, index):
        return self.data[index]

您将能够编写如下代码：

chromosome1 = Chromosome("chromosome1", [
    Gene('geneA', [10, 20]), 
    Gene('geneB', [12, 45]), 
    Gene('geneC', [36, 50])
])

并且认为如下：

print(chromosome1.name) # Print the chromosome name
>>> chromosome1
print(chromosome1[0].name) # The name of the first gene
>>> geneA
print(chromosome1[1].name) # The name of the second gene
>>> geneB
print(chromosome1[0][1])  # The second value of the first gene
20

你也可以有一个choromosomes列表（这实际上是你想要的）：

lchrom = [chromosome1, ...]
print(lchrom[0][1]) # The second gene of the first choromosome in the list. (geneB)
print(lchrom[0][1][0]) # The first value second gene of the first choromosome in the list. (12)

Python变量和索引

1 个答案:

POO方法