Question

我有一个如下所示的列表：

mylist = ['Probes', 'Gene.symbol', 'Gene.Title', 'GO1', 'GO2', 'GO3', 'ADX_KD_06.ip', 'ADX_KD_24.ip', 'ADX_LG_06.ip', 'ADX_LG_24.ip', 'ADX_LV_06.ip', 'ADX_LV_24.ip', 'ADX_SP_06.ip', 'ADX_SP_24.ip', 'ADX_LN_06.id', 'ALM_LN_06.id', 'ALM_LV_06.ip', 'ALM_SP_06.ip', 'K3SPG_LV_06.ip', 'K3SPG_SP_06.ip', 'KKK_LN_06.id', 'KKK_LV_06.ip', 'KKK_SP_06.ip', 'ENDCN_LV_06.in', 'ENDCN_SP_06.in', 'bCD_LV_06.ip', 'bCD_SP_06.ip', 'ADX_LV_06.id', 'ADX_SP_06.id', 'ALM_LV_06.id', 'ALM_SP_06.id', 'D35_LN_06.id', 'K3SPG_LN_06.id', 'K3_LV_06.id', 'K3_SP_06.id', 'bCD_LN_06.id', 'D35_LV_06.id', 'D35_SP_06.id', 'K3SPG_LV_06.id', 'K3SPG_SP_06.id', 'bCD_LV_06.id', 'bCD_SP_06.id', 'ENDCN_KD_06.in', 'ENDCN_LG_06.in', 'Probes', 'Gene.symbol', 'ADX_KD_06.ip', 'ADX_KD_24.ip', 'ADX_LG_06.ip', 'ADX_LG_24.ip', 'ADX_LV_06.ip', 'ADX_LV_24.ip', 'ADX_SP_06.ip', 'ADX_SP_24.ip', 'ADX_LN_06.id', 'ALM_LN_06.id', 'ALM_LV_06.ip', 'ALM_SP_06.ip', 'K3SPG_LV_06.ip', 'K3SPG_SP_06.ip', 'KKK_LN_06.id', 'KKK_LV_06.ip', 'KKK_SP_06.ip', 'ENDCN_LV_06.in', 'ENDCN_SP_06.in', 'bCD_LV_06.ip', 'bCD_SP_06.ip', 'ADX_LV_06.id', 'ADX_SP_06.id', 'ALM_LV_06.id', 'ALM_SP_06.id', 'D35_LN_06.id', 'K3SPG_LN_06.id', 'K3_LV_06.id', 'K3_SP_06.id', 'bCD_LN_06.id', 'D35_LV_06.id', 'D35_SP_06.id', 'K3SPG_LV_06.id', 'K3SPG_SP_06.id', 'bCD_LV_06.id', 'bCD_SP_06.id', 'ENDCN_KD_06.in', 'ENDCN_LG_06.in']
# index =    0,        1,             2,          3,      4,    5,      6, ....

我们可以看到上面的列表分为两部分，由'Probes', 'Gene.symbol', 'Gene.Title', 'GO1', 'GO2', 'GO3'和'Probes', 'Gene.symbol'分隔。

第一部分和第二部分的正则表达式是：

([\w\d]+)_(\w\w)_(\d\d)\.(\w\w)
  rg1      rg2     rg3    rg4

哪个匹配字符串，例如ADX_SP_06.ip或K3SPG_LN_06.id

我想要做的是创建一个由rg1 + rg3形成的密钥的字典其中值存储mylist的第一和第二部分中成员的索引。结果将如下所示。

first_part_dict = {
      'bCD06'    : { 25, 26, 35, 40, 41 },
      'ADX06'    : { 6,  8,  10, 12, 14, 27, 28 },
      'ADX24'    : { 7, 9, 11, 13 },
      'ALM06'    : { 15, 16, 17, 29, 30 },
      'K3SPG06'  : { 18, 19, 32, 38, 39 },
      'KKK06'    : { 20, 21, 22 },
      'K306'     : { 33, 34},
      'ENDCN06'  : { 23, 24, 42,43 },
      'D3506'    : { 31, 36, 37 },
}

second_part_dict = {
      'bCD06'    : { 65, 66, 75, 80, 81 },
      'ADX06'    : { 46, 48, 50, 52, 54, 67, 68},
      'ADX24'    : { 47, 49, 51, 53 },
      'ALM06'    : { 55, 56, 57, 69, 70 },
      'K3SPG06'  : { 58, 59, 72, 78, 79 },
      'KKK06'    : { 60, 61, 62 },
      'K306'     : { 73, 74},
      'ENDCN06'  : { 63, 64, 82, 83},
      'D3506'    : { 71, 76, 77 },
}

如何使用Python实现这一目标？

Answer 1

这不是很优雅，但它应该完成工作。我根据您的预期输出假设您认为密钥为rg1 + rg3，而不是rg1 + rg2。

import re, collections

mylist = ['Probes', 'Gene.symbol', 'Gene.Title', 'GO1', 'GO2', 'GO3', 'ADX_KD_06.ip', 'ADX_KD_24.ip', 'ADX_LG_06.ip', 'ADX_LG_24.ip', 'ADX_LV_06.ip', 'ADX_LV_24.ip', 'ADX_SP_06.ip', 'ADX_SP_24.ip', 'ADX_LN_06.id', 'ALM_LN_06.id', 'ALM_LV_06.ip', 'ALM_SP_06.ip', 'K3SPG_LV_06.ip', 'K3SPG_SP_06.ip', 'KKK_LN_06.id', 'KKK_LV_06.ip', 'KKK_SP_06.ip', 'ENDCN_LV_06.in', 'ENDCN_SP_06.in', 'bCD_LV_06.ip', 'bCD_SP_06.ip', 'ADX_LV_06.id', 'ADX_SP_06.id', 'ALM_LV_06.id', 'ALM_SP_06.id', 'D35_LN_06.id', 'K3SPG_LN_06.id', 'K3_LV_06.id', 'K3_SP_06.id', 'bCD_LN_06.id', 'D35_LV_06.id', 'D35_SP_06.id', 'K3SPG_LV_06.id', 'K3SPG_SP_06.id', 'bCD_LV_06.id', 'bCD_SP_06.id', 'ENDCN_KD_06.in', 'ENDCN_LG_06.in', 'Probes', 'Gene.symbol', 'ADX_KD_06.ip', 'ADX_KD_24.ip', 'ADX_LG_06.ip', 'ADX_LG_24.ip', 'ADX_LV_06.ip', 'ADX_LV_24.ip', 'ADX_SP_06.ip', 'ADX_SP_24.ip', 'ADX_LN_06.id', 'ALM_LN_06.id', 'ALM_LV_06.ip', 'ALM_SP_06.ip', 'K3SPG_LV_06.ip', 'K3SPG_SP_06.ip', 'KKK_LN_06.id', 'KKK_LV_06.ip', 'KKK_SP_06.ip', 'ENDCN_LV_06.in', 'ENDCN_SP_06.in', 'bCD_LV_06.ip', 'bCD_SP_06.ip', 'ADX_LV_06.id', 'ADX_SP_06.id', 'ALM_LV_06.id', 'ALM_SP_06.id', 'D35_LN_06.id', 'K3SPG_LN_06.id', 'K3_LV_06.id', 'K3_SP_06.id', 'bCD_LN_06.id', 'D35_LV_06.id', 'D35_SP_06.id', 'K3SPG_LV_06.id', 'K3SPG_SP_06.id', 'bCD_LV_06.id', 'bCD_SP_06.id', 'ENDCN_KD_06.in', 'ENDCN_LG_06.in']

regex = re.compile(r'([\w\d]+)_(\w\w)_(\d\d)\.(\w\w)')
first_part_dict = collections.defaultdict(list)
second_part_dict = collections.defaultdict(list)
# second instance of 'Probes', to separate the first and second parts
cutoff_index = mylist.index('Probes', 1) 

for i, string in enumerate(mylist):
    matched = regex.match(string)
    if not matched:
        continue
    rg1, rg2, rg3, rg4 = matched.groups()
    key = rg1 + rg3
    if i < cutoff_index:
        first_part_dict[key].append(i)
    else:
        second_part_dict[key].append(i)

结果：

>>> first_part_dict
defaultdict(<class 'list'>, {'ALM06': [15, 16, 17, 29, 30], 'K3SPG06': [18, 19, 32, 38, 39], 'bCD06': [25, 26, 35, 40, 41], 'ADX24': [7, 9, 11, 13], 'ENDCN06': [23, 24, 42, 43], 'KKK06': [20, 21, 22], 'K306': [33, 34], 'ADX06': [6, 8, 10, 12, 14, 27, 28], 'D3506': [31, 36, 37]})
>>> second_part_dict
defaultdict(<class 'list'>, {'ALM06': [55, 56, 57, 69, 70], 'K3SPG06': [58, 59, 72, 78, 79], 'bCD06': [65, 66, 75, 80, 81], 'ADX24': [47, 49, 51, 53], 'ENDCN06': [63, 64, 82, 83], 'KKK06': [60, 61, 62], 'K306': [73, 74], 'ADX06': [46, 48, 50, 52, 54, 67, 68], 'D3506': [71, 76, 77]})

基于正则表达式将列表的结构化元素索引转换为字典

1 个答案: