如何在python列表中包含许多类别组合(72个变量)

时间:2014-06-27 19:49:33

标签: python

我将数据存储在如下组织的列表列表中:

lst = [
      ['FHControl', G, A]
      ['MNHDosed', G, C]
      ]

对于lst:row [0]中的行,总共有12个类别(我在上面的示例代码中列出了两个类别)。对于行[1]和行[2],我只关注这些字母的6个组合。因此,我在lst中每行有72种可能的数据组合,需要计算每个组合的实例,而不必编写几十个嵌套的if循环。

我正在尝试创建两个函数来解析这些列表并对这72个组合的发生率进行分析。我怎样才能使用两个函数,比如我下面开始编写的更新这些变量的函数?我是否需要将字典构造为类变量,以便在迭代这两个函数时可以继续更新它们?任何指导都会很棒!

这是我目前将所有72个变量初始化为6个字典的代码(对于行[1]和行[2]中的6个字母组合):

def baseparser(lst):
    TEMP = dict.fromkeys('FHDosed FHControl FNHDosed FNHControl '
                         'FTDosed FTControl MHDosed MHControl '
                         'MNHDosed MNHControl MTDosed MTControl'.split(), 0)
    TRI_1, TRI_2, TRV_1, TRV_2, TRV_3, TRV_4 = ([dict(TEMP) for i in range(6)])

    for row in lst:
        if row[0] == 'FHDosed':
            binner(row[0], row[1], row[2])
        if row[0] == 'FHControl':
            binner(row[0], row[1], row[2])
        etc.

def binner(key, q, s):
    if (q == 'G' and s == 'A') or (q =='C' and s =='T'):
        TRI_1[key] += 1
    elif (q == 'A' and s == 'G') or (q =='T' and s =='C'):
        TRI_2[key] += 1
    elif (q == 'G' and s == 'T') or (q =='C' and s =='A'):
        TRV_1[key] += 1
    elif (q == 'G' and s == 'C') or (q =='C' and s =='G'):
        TRV_1[key] += 1
    elif (q == 'A' and s == 'T') or (q =='T' and s =='A'):
        TRV_1[key] += 1
    elif (q == 'A' and s == 'C') or (q =='T' and s =='G'):
        TRV_1[key] += 1

1 个答案:

答案 0 :(得分:4)

您的代码可以简化为:

TEMP = dict.fromkeys('''FHDosed FHControl FNHDosed FNHControl FTDosed FTControl MHDosed 
                      MHControl MNHDosed MNHControl MTDosed MTControl'''.split(), 0)
TRI_1, TRI_2, TRV_1, TRV_2, TRV_3, TRV_4 = [TEMP.copy() for i in range(6)]

dmap = {
    ('G', 'A'): TRI_1,
    ('C', 'T'): TRI_1,
    ('A', 'G'): TRI_2,
    ('T', 'C'): TRI_2,        
    ('G', 'C'): TRV_1,
    ('C', 'G'): TRV_1,        
    ('A', 'T'): TRV_1,
    ('T', 'A'): TRV_1,        
    }

for row in lst:
    key, q, s = row
    dmap[q, s][key] += 1

另一种可能性是使用一个dicts字典而不是6个dicts:

TEMP = dict.fromkeys('''FHDosed FHControl FNHDosed FNHControl FTDosed FTControl MHDosed 
                      MHControl MNHDosed MNHControl MTDosed MTControl'''.split(), 0)
TR = {key:TEMP.copy() for key in ('TRI_1', 'TRI_2', 'TRV_1', 'TRV_2', 'TRV_3', 'TRV_4')}


dmap = {
    ('G', 'A'): 'TRI_1',
    ('C', 'T'): 'TRI_1',
    ('A', 'G'): 'TRI_2',
    ('T', 'C'): 'TRI_2', 
    ('G', 'C'): 'TRV_1',
    ('C', 'G'): 'TRV_1', 
    ('A', 'T'): 'TRV_1',
    ('T', 'A'): 'TRV_1',        
    }

lst = [
      ['FHControl', 'G', 'A'],
      ['MNHDosed', 'G', 'C']
      ]

for row in lst:
    key, q, s = row
    TR[dmap[q, s]][key] += 1

print(TR)

这样做的好处是你的命名空间中的dicts更少,并且稍后使用dicts的dict而不是硬编码6 dicts来重构代码可能更容易。


跟进Midnighter的建议,如果你有pandas,你可以用一个DataFrame替换dicts的dict。然后可以使用pd.crosstabs计算对的频率,如下所示:

import pandas as pd

dmap = {
    'GA': 'TRI_1',
    'CT': 'TRI_1',
    'AG': 'TRI_2',
    'TC': 'TRI_2', 
    'GC': 'TRV_1',
    'CG': 'TRV_1', 
    'AT': 'TRV_1',
    'TA': 'TRV_1',        
    }

lst = [
      ['FHControl', 'G', 'A'],
      ['MNHDosed', 'G', 'C']
      ]

df = pd.DataFrame(lst, columns=['key', 'q', 's'])
df['tr'] = (df['q']+df['s']).map(dmap)

print(df)
#          key  q  s     tr
# 0  FHControl  G  A  TRI_1
# 1   MNHDosed  G  C  TRV_1

print(pd.crosstab(rows=[df['key']], cols=[df['tr']]))

产量

tr         TRI_1  TRV_1
key                    
FHControl      1      0
MNHDosed       0      1