我在网络中有需求的以下文本文件。
Origin 1
1 : 0.0; 2 : 100.0; 3 : 100.0; 4 : 500.0; 5 : 200.0;
6 : 300.0; 7 : 500.0; 8 : 800.0; 9 : 500.0; 10 : 1300.0;
11 : 500.0; 12 : 200.0; 13 : 500.0; 14 : 300.0; 15 : 500.0;
16 : 500.0; 17 : 400.0; 18 : 100.0; 19 : 300.0; 20 : 300.0;
21 : 100.0; 22 : 400.0; 23 : 300.0; 24 : 100.0;
Origin 2
1 : 100.0; 2 : 0.0; 3 : 100.0; 4 : 200.0; 5 : 100.0;
6 : 400.0; 7 : 200.0; 8 : 400.0; 9 : 200.0; 10 : 600.0;
11 : 200.0; 12 : 100.0; 13 : 300.0; 14 : 100.0; 15 : 100.0;
16 : 400.0; 17 : 200.0; 18 : 0.0; 19 : 100.0; 20 : 100.0;
21 : 0.0; 22 : 100.0; 23 : 0.0; 24 : 0.0;
Origin 3
1 : 100.0; 2 : 100.0; 3 : 0.0; 4 : 200.0; 5 : 100.0;
6 : 300.0; 7 : 100.0; 8 : 200.0; 9 : 100.0; 10 : 300.0;
11 : 300.0; 12 : 200.0; 13 : 100.0; 14 : 100.0; 15 : 100.0;
16 : 200.0; 17 : 100.0; 18 : 0.0; 19 : 0.0; 20 : 0.0;
21 : 0.0; 22 : 100.0; 23 : 100.0; 24 : 0.0;
... records 4-23 elided ...
Origin 24
1 : 100.0; 2 : 0.0; 3 : 0.0; 4 : 200.0; 5 : 0.0;
6 : 100.0; 7 : 100.0; 8 : 200.0; 9 : 200.0; 10 : 800.0;
11 : 600.0; 12 : 500.0; 13 : 700.0; 14 : 400.0; 15 : 400.0;
16 : 300.0; 17 : 300.0; 18 : 0.0; 19 : 100.0; 20 : 400.0;
21 : 500.0; 22 : 1100.0; 23 : 700.0; 24 : 0.0;
现在我需要创建一个字典，它应该类似于：
{(1,1):0.0, (1,2):100.0, (1, 3):100.0, .......
(2, 1):100.0, (2,2):0, ......}
元组元素例如(1, 2)代表原点和目的地，值代表需求（100.0密钥为(1, 2)。）
我尝试了以下内容：
with open("trips.txt", "r") as f:
line = f.readline()
line = f.readline()
ind = 0
while len(line):
line = line.strip(';')
l = line.split()
print l
ind = ind + 1
if(ind == 5):
line = f.readline()
line = f.readline()
line = f.readline()
ind = 0
node = node + 1
else:
line = f.readline()
但是我认为我不会去任何地方......
答案 0 :(得分：3)
你绝对不会去任何地方，因为你根本没有对字典做过任何参考。
我将在此为您概述一个流程;你能填写详细资料吗？
my_dict = {}
while not EOF:
# read the "Origin" line
line = f.readline()
# extract the number on the right
origin_num = int( line.split()[-1] )
# Read the data lines
for _ in range(5): # each data chunk has 5 lines
data_line = readline()
entries = data_line.split(';') # split at semicolons
for field in entries:
y_key, value = field.split(:)
# Now, you need to convert both of those to integers,
# combine v_key with the origin_num,
# and insert that value into my_dict.
这会让你感动吗？ 请注意，您还需要处理空行，检测文件结尾等。
答案 1 :(得分：2)
嗯，如果要提取数据，则需要逐行解析，算法大致应该是：
if空行，跳过if该行以＆Origin;＆＃39;起始＆＃39;捕获它之后的数字（origin_no）else用分号和每个元素拆分行：
element_no）value_no）(origin_no, element_no): value_no 实施起来非常简单：
result = {} # we'll store our result in this dict
origin_no = 0 # our starting Origin number in case the file doesn't begin with one
with open("trips.txt", "r") as f:
for line in f:
line = line.rstrip() # we're not interested in the newline at the end
if not line: # empty line, skip
continue
if line.startswith("Origin"):
origin_no = int(line[7:].strip()) # grab the integer following Origin
else:
elements = line.split(";") # get our elements by splitting by semi-colon
for element in elements: # loop through each of them:
if not element: # we're not interested in the last element
continue
element_no, element_value = element.split(":") # get our pair
# beware, these two are now most likely padded strings!
# that's why we'll strip them from whitespace and convert to integer/float
result[(origin_no, int(element_no.strip()))] = float(element_value.strip())
# Done!
答案 2 :(得分：1)
您可以尝试：
with open('trips.txt', 'r') as f:
dic = {}
try:
while True:
num = int(f.next().split()[1])
lst = []
for _ in xrange(5):
lst.append(f.next().strip().split(';'))
f.next()
for n in lst:
for l in n:
if l != '':
tmp = l.strip().split(':')
dic[(num, int(tmp[0]))] = float(tmp[1])
except StopIteration:
print dic
输出：
{(1, 21): 100.0, (1, 3): 100.0, (2, 18): 0.0, (2, 8): 400.0, (1, 17): 400.0, (2, 1): 100.0, (1, 15): 500.0, (2, 22): 100.0....etc}
答案 3 :(得分：1)
另一种方法 -
nw.usage是一个具有使用内容的文件..
正如我在下面的代码中所评论的那样..如果你想维护插入顺序，请使用collections.OrderedDict（）。
希望它有所帮助！
#!/usr/bin/env python
import re
#import collections
with open('nw.usage', 'r') as f:
usage_dict = {}
#Use collections.OrderedDict() if you want to maintain insertion order
origin_val = ''
for line in f:
if re.search('Origin', line):
origin_val = line.rstrip()[-1]
else:
hr_demand = line.strip().split(';')
for hr in hr_demand:
if not hr:
continue
hour = hr.split(':')[0].strip()
usage = hr.split(':')[1].strip()
usage_dict[(origin_val, hour)] = usage
print usage_dict
输出是 -
{('1', '17'): '400.0', ('2', '2'): '0.0', ('2', '17'): '200.0', ('1', '20'): '300.0', ('1', '18'): '100.0', ('2', '20'): '100.0', ('1', '13'): '500.0', ('1', '6'): '300.0', ('2', '13'): '300.0', ('1', '24'): '100.0', ('2', '7'): '200.0', ('2', '24'): '0.0', ('1', '2'): '100.0', ('1', '16'): '500.0', ('2', '3'): '100.0', ('2', '18'): '0.0', ('1', '21'): '100.0', ('2', '23'): '0.0', ('1', '12'): '200.0', ('2', '14'): '100.0', ('2', '8'): '400.0', ('1', '5'): '200.0', ('2', '10'): '600.0', ('2', '4'): '200.0', ('2', '19'): '100.0', ('1', '22'): '400.0', ('1', '1'): '0.0', ('2', '22'): '100.0', ('1', '15'): '500.0', ('2', '15'): '100.0', ('2', '9'): '200.0', ('1', '11'): '500.0', ('1', '4'): '500.0', ('2', '11'): '200.0', ('1', '9'): '500.0', ('2', '5'): '100.0', ('1', '23'): '300.0', ('1', '14'): '300.0', ('2', '1'): '100.0', ('2', '16'): '400.0', ('1', '19'): '300.0', ('2', '21'): '0.0', ('1', '10'): '1300.0', ('1', '7'): '500.0', ('2', '12'): '100.0', ('1', '8'): '800.0', ('2', '6'): '400.0', ('1', '3'): '100.0'}