Question

12245933,1418,1
12245933,1475,2
134514060,6112,3
134514064,10096,4
12245933,1536,5
...
134514097,16200,38
12245933,1475,39

我想知道每row[0] row[1]

中相同值重新出现的距离

12245933  has the value 1475 in line 39 and line 2 ..
i want to know all the possible occurrences of 1475 for 12245933 in a file.

例如：

#datafile parser
def parse_data(file):
    pc_elements = defaultdict(list)
    addr_elements = defaultdict(list)
    with open(file, 'rb') as f:
        line_number = 0
        csvin = csv.reader((x.replace('\0','') for x in f), delimiter = ',')
        for row in csvin:
            try:
                pc_elements[int(row[0])].append(line_number)
                addr_elemets[int(row[1])].append(line_number)
                line_number += 1
            except:
                print row
                line_number += 1
                pass

我试过的代码。

{{1}}

也许我们可以在pc_elements dict中添加row [1]？并从中获取索引？

Answer 1

使用tuple s作为词典键：

In [63]: d='''
    ...: 12245933,1418,1
    ...: 12245933,1475,2
    ...: 134514060,6112,3
    ...: 134514064,10096,4
    ...: 12245933,1536,5
    ...: 134514097,16200,38
    ...: 12245933,1475,39
    ...: '''

In [64]: from collections import defaultdict
    ...: dic=defaultdict(list)
    ...: for l in d.split():
    ...:     tup=tuple(int(i) for i in l.split(','))
    ...:     dic[tup[:2]].append(tup[2])

In [65]: dic[(12245933, 1475)]
Out[65]: [2, 39]

Answer 2

使用嵌套词典。将1224953映射到一个字典，该字典将1475映射到值出现的行号列表。

所以你的最后一本字典看起来像{1224953 =＆gt; {1475 =＆gt; [39,2]}}

与多个键的词典在Python

2 个答案: