将文本元素保存到python中的词典

我有一个以下格式的文本文件：

attr1    1,3,7,6,8,12,24,56
attr2    1,2,3
attr4    56,45,48,23,24,25,29,90,56,57,58,59
attr5    1,2,3,45,6,7,8,9,34,33

我想创建一个dict，其中数字是键，如果包含attr，那么每个键都必须包含在一个列表中。为了更具体地说明我所写的例子，dict必须是：

    1: [attr1,attr2,attr5]
    2: [attr2,attr5]
    3: [attr1,attr2,attr5]
    6: [attr1, attr5] 
etc...

我试图实现它，并编写了以下代码，但它不起作用。这是我的密码：

file2 = open("attrs.txt","r")
lines2 = file2.readlines()
d = dict()
list1 = []
for x in lines2:
    x = x.strip()
    x = x.split('\t')
    y = x[0]
    list1.append(x[1].split(','))
    for i in list1:
        d[i] = y

---

您可以使用集合。defaultdict：

import collections
import re
file_data = [[a, list(map(int, b.split(',')))] for a, b in [re.split('\s+', i.strip('\n')) for i in open('filename.txt')]]
d = collections.defaultdict(list)
for a, b in file_data:
  for i in b:
    d[i].append(a)

print(dict(d))

输出：

{1: ['attr1', 'attr2', 'attr5'], 2: ['attr2', 'attr5'], 3: ['attr1', 'attr2', 'attr5'], 6: ['attr1', 'attr5'], 7: ['attr1', 'attr5'], 8: ['attr1', 'attr5'], 9: ['attr5'], 12: ['attr1'], 23: ['attr4'], 24: ['attr1', 'attr4'], 25: ['attr4'], 90: ['attr4'], 29: ['attr4'], 33: ['attr5'], 34: ['attr5'], 45: ['attr4', 'attr5'], 48: ['attr4'], 56: ['attr1', 'attr4', 'attr4'], 57: ['attr4'], 58: ['attr4'], 59: ['attr4']}

{1: ['attr1', 'attr2', 'attr5'], 2: ['attr2', 'attr5'], 3: ['attr1', 'attr2', 'attr5'], 33: ['attr5'], 6: ['attr1', 'attr5'], 7: ['attr1', 'attr5'], 8: ['attr1', 'attr5'], 9: ['attr5'], 12: ['attr1'], 34: ['attr5'], 45: ['attr4', 'attr5'], 48: ['attr4'], 56: ['attr1', 'attr4', 'attr4'], 90: ['attr4'], 57: ['attr4'], 23: ['attr4'], 24: ['attr1', 'attr4'], 25: ['attr4'], 58: ['attr4'], 59: ['attr4'], 29: ['attr4']}

或者使用

itertools.groupby

的更短但更复杂的解决方案：

import itertools
new_data = list(itertools.chain(*[[[i, a] for i in b] for a, b in file_data]))
final_result = {a:[b for _, b in c] for a, c in itertools.groupby(sorted(new_data, key=lambda x:x[0]), key=lambda x:x[0])}

输出：

{1: ['attr1', 'attr2', 'attr5'], 2: ['attr2', 'attr5'], 3: ['attr1', 'attr2', 'attr5'], 6: ['attr1', 'attr5'], 7: ['attr1', 'attr5'], 8: ['attr1', 'attr5'], 9: ['attr5'], 12: ['attr1'], 23: ['attr4'], 24: ['attr1', 'attr4'], 25: ['attr4'], 90: ['attr4'], 29: ['attr4'], 33: ['attr5'], 34: ['attr5'], 45: ['attr4', 'attr5'], 48: ['attr4'], 56: ['attr1', 'attr4', 'attr4'], 57: ['attr4'], 58: ['attr4'], 59: ['attr4']}

{1: ['attr1', 'attr2', 'attr5'], 2: ['attr2', 'attr5'], 3: ['attr1', 'attr2', 'attr5'], 33: ['attr5'], 6: ['attr1', 'attr5'], 7: ['attr1', 'attr5'], 8: ['attr1', 'attr5'], 9: ['attr5'], 12: ['attr1'], 34: ['attr5'], 45: ['attr4', 'attr5'], 48: ['attr4'], 56: ['attr1', 'attr4', 'attr4'], 90: ['attr4'], 57: ['attr4'], 23: ['attr4'], 24: ['attr1', 'attr4'], 25: ['attr4'], 58: ['attr4'], 59: ['attr4'], 29: ['attr4']}

---

如果您乐于使用第三方库，

pandas

提供了一种方法：

import pandas as pd
from io import StringIO

mystr = StringIO("""
attr1    1,3,7,6,8,12,24,56
attr2    1,2,3
attr4    56,45,48,23,24,25,29,90,56,57,58,59
attr5    1,2,3,45,6,7,8,9,34,33""")

# replace mystr with 'file.csv'
df = pd.read_csv(mystr, delim_whitespace=True, header=None, names=['attrs', 'lists'])

# convert identifier column to int
df['attrs'] = df['attrs'].str[4:].map(int)

# split and convert attrs to int
df['lists'] = [list(map(int, x.split(','))) for x in df['lists']]

d = df.set_index('attrs')['lists'].to_dict()

# {1: [1, 3, 7, 6, 8, 12, 24, 56],
#  2: [1, 2, 3],
#  4: [56, 45, 48, 23, 24, 25, 29, 90, 56, 57, 58, 59],
#  5: [1, 2, 3, 45, 6, 7, 8, 9, 34, 33]}

---

您可以使用本机Python函数来实现这一点

我为每个字典值使用集合，并最终将它们转换为排序列表，以防您的数据包含重复的条目

keys = set()
d = dict()
f = open("attrs.txt","r")
for line in f:
    attr,newkeys = line.strip().split()
    newkeys = [int(x) for x in newkeys.split(',')]
    for key in newkeys:
        if key not in keys:
            d[key] = set()
            keys.add(key)
        d[key].add(attr)
for key in list(d):
    d[key]=sorted(list(d[key]))
f.close()

---

这取决于您的目的以及实际数据格式的复杂程度。对于包含更多列的输入文件，或者在文件读取过程中没有其他需要完成的目标，Pandas（如@jpp所做的）将是更好的方法。

使用

defaultdict

和

设置

：

#!/usr/bin/env python3

import io
import collections
from pprint import pprint


fdata = """attr1    1,3,7,6,8,12,24,56
attr2   1,2,3
attr4   56,45,48,23,24,25,29,90,56,57,58,59
attr5   1,2,3,45,6,7,8,9,34,33
"""

# with open('attrs.txt') as f:
with io.StringIO(fdata) as f:
    d = collections.defaultdict(set)
    for line in f:
        name, keys = line.strip().split()
        for k in keys.split(','):
            d[int(k)].add(name)

pprint(d)

---

您也可以这样做：

#
# 1) parse the input into an array of dictionaries 
#    having a one attribute list for each key
# 
attribKeys = [ dict.fromkeys(ak[1].split(","),[ak[0]]) for ak in [ line.split("\t") for line in line2.split("\n")] ]

#
# 2) merge the dictionaries concatenating lists for each key
#
import functools as fn
d = fn.reduce(lambda d,a: dict(list(d.items()) + [(k,d.get(k,[])+v) for k,v in a.items()]),attribKeys)

# d will contain: 
#
# {'1': ['attr1', 'attr2', 'attr5'], 
#  '2': ['attr2', 'attr5'], 
#  '3': ['attr1', 'attr2', 'attr5'],
#  '6': ['attr1', 'attr5'], 
#  '7': ['attr1', 'attr5'], 
#  '8': ['attr1', 'attr5'], 
#  '9': ['attr5'],
#  '12': ['attr1'], 
#  '23': ['attr4'], 
#  '24': ['attr1', 'attr4'],
#  '25': ['attr4'], 
#  '29': ['attr4'],
#  '33': ['attr5'], 
#  '34': ['attr5'], 
#  '45': ['attr4', 'attr5'], 
#  '48': ['attr4'], 
#  '56': ['attr1', 'attr4'], 
#  '57': ['attr4'], 
#  '58': ['attr4'], 
#  '59': ['attr4'], 
#  '90': ['attr4']}

---

你的问题是列表1…我尝试了你的代码，但我得到了这个错误：

ValueError:太多的值无法解包（预期为2）

@LeeYaan奇怪，你确定你问题中发布的数据正是你文件中数据的格式吗？@LeeYaan奇怪，当我从文件中读取它时，它对我有效。哪个版本会引发错误？

collections.defaultdict

解决方案，还是使用

itertools

的解决方案？