12245933,1418,1
12245933,1475,2
134514060,6112,3
134514064,10096,4
12245933,1536,5
...
134514097,16200,38
12245933,1475,39
我想知道对于每个 row[0]
,相同值在 row[1]
中再次出现的距离
例如:
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.
我试过的代码。
#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
也许我们也可以在 pc_elements 字典中添加 row[1]?并从中获取索引?
最佳答案
使用元组
作为你的字典键:
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]
关于python - python中具有多个键的字典,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22086725/