Question

所以我有多行看起来像这样的文件（空格分隔符文件）：

A1BG      P04217     VAR_018369  p.His52Arg     Polymorphism  rs893184    -
A1BG      P04217     VAR_018370  p.His395Arg    Polymorphism  rs2241788   -
AAAS      Q9NRG9     VAR_012804  p.Gln15Lys     Disease       -           Achalasia

如何使字典在第二列中查找id并将数字（在单词之间）存储在第四列。

我尝试了这个，但它给了我超出范围的索引

lookup = defaultdict(list)
with open ('humsavar.txt', 'r') as humsavarTxt:
    for line in csv.reader(humsavarTxt):
        code = re.match('[a-z](\d+)[a-z]', line[1], re.I)
        if code: 
            lookup[line[-2]].append(code.group(1))

print lookup['P04217']

Answer 1

以下是原始代码的变体：

import csv, re
from collections import defaultdict

lookup = defaultdict(list)
with open('humsavar.txt', 'rb') as humsavarTxt:
    reader = csv.reader(humsavarTxt, delimiter=" ", skipinitialspace=True)
    for line in reader:
        code = re.search(r'(\d+)', line[3])
        lookup[line[1]].append(int(code.group(1)))

产生

>>> lookup
defaultdict(<type 'list'>, {'P04217': [52, 395], 'Q9NRG9': [15]})
>>> lookup['P04217']
[52, 395]

Answer 2

如果id和数字始终位于第二和第四列，并且它始终以空格分隔，则不需要使用常规表达式。你可以在空格上拆分：

lookup = defaultdict(list)
with open ('humsavar.txt', 'r') as humsavarTxt:
    for line in humsavarTxt:
         lookup[line.split(' ')[1]].append(line.split(' ')[3])

Answer 3

如果你想要一个纯字典，这可行：

d={}
with open(your_file,'rb') as f:
    for line in f:
        l=line.split()
        num=int(re.search(r'(\d+)',l[3]).group(1))
        d.setdefault(l[1],[]).append(num)

打印：

{'P04217': [52, 395], 'Q9NRG9': [15]}

对于非正则表达式解决方案，您也可以这样做：

d={}
with open(your_file,'rb') as f:
    for line in f:
        els=line.split()
        num=int(''.join(c for c in els[3] if c.isdigit()))
        d.setdefault(els[1],[]).append(num)

在python中查找字典

3 个答案: