python - 如何通过 Python 中的 for 循环传递列表列表？

Question

我有一个列表列表:

sample = [['TTTT', 'CCCZ'], ['ATTA', 'CZZC']]
count = [[4,3],[4,2]]
correctionfactor  = [[1.33, 1.5],[1.33,2]]

我计算每个字符 (pi) 的频率，对其求平方然后求和(然后我计算 het = 1 - 求和)。

The desired output [[1,2],[1,2]] #NOTE: This is NOT the real values of expected output. I just need the real values to be in this format. 

问题:我不知道如何在此循环中传递列表列表(样本，计数)以提取所需的值。我以前使用此代码仅传递了一个列表(例如 ['TACT','TTTT'..])。

• 我怀疑我需要添加一个更大的 for 循环，它索引样本中的每个元素(即索引 sample[0] = ['TTTT', 'CCCZ'] 和 sample[1] = ['ATTA', 'CZZC']。我不确定如何将其合并到代码中。

**代码

list_of_hets = []
for idx, element in enumerate(sample):
    count_dict = {}
    square_dict = {}
    for base in list(element):
         if base in count_dict:
            count_dict[base] += 1
        else:
            count_dict[base] = 1
    for allele in count_dict: #Calculate frequency of every character
        square_freq = (count_dict[allele] / count[idx])**2 #Square the frequencies
        square_dict[allele] = square_freq        
    pf = 0.0
    for i in square_dict:
        pf += square_dict[i]   # pf --> pi^2 + pj^2...pn^2 #Sum the frequencies
    het = 1-pf                    
    list_of_hets.append(het)
print list_of_hets

"Failed" OUTPUT:
line 70, in <module>
square_freq = (count_dict[allele] / count[idx])**2
TypeError: unsupported operand type(s) for /: 'int' and 'list'er

Answer 1

我不完全清楚您想如何处理数据中的“Z”项，但这段代码复制了 https://eval.in/658468 中示例数据的输出。

from __future__ import division

bases = set('ACGT')
#sample = [['TTTT', 'CCCZ'], ['ATTA', 'CZZC']]
sample = [['ATTA', 'TTGA'], ['TTCA', 'TTTA']]

list_of_hets = []
for element in sample:
    hets = []
    for seq in element:
        count_dict = {}
        for base in seq:
            if base in count_dict:
                count_dict[base] += 1
            else:
                count_dict[base] = 1
        print count_dict

        #Calculate frequency of every character
        count = sum(1 for u in seq if u in bases)
        pf = sum((base / count) ** 2 for base in count_dict.values())
        hets.append(1 - pf)
    list_of_hets.append(hets)

print list_of_hets

输出

{'A': 2, 'T': 2}
{'A': 1, 'T': 2, 'G': 1}
{'A': 1, 'C': 1, 'T': 2}
{'A': 1, 'T': 3}
[[0.5, 0.625], [0.625, 0.375]]

通过使用 collections.Counter 而不是 count_dict 可以进一步简化此代码。

顺便说一句，如果不在“ACGT”中的符号总是“Z”，那么我们可以加快count 计算。摆脱 bases = set('ACGT') 并改变

count = sum(1 for u in seq if u in bases)

到

count = sum(1 for u in seq if u != 'Z')