从文件到字典的Python列表

Question

我有一个笔记文件，我试图转换成字典。我得到的脚本工作，但未能输出数据我寻找时，有重复的值。

简单地说，就是下面用#隔开的文件命令或注释。我取这个列表，把第一列的键分开通过#，其余部分是注释或定义。然后我检查我要查找的神奇单词，解析它，匹配它，然后输出。

卡片 文件如下

> car # automobile 4 wheels and run 
> washington dc # the capital of United States 
> fedora # an operating distro
> cat file  # reads the file 
> car nissan # altima
> car nissan # altima ## first car 
> car nissan # maxima
> car nissan # rougue

flashcard_dict = dict()
flashcard_file = open('FlashCards','r')
enter = input("Searching nemo: ")

firstcolumn_str_list = list()
for x in flashcard_file:
flashcard_sprint = x.strip()
flascard_clean = flashcard_sprint.split("#",1)
firstcolumn_str = flascard_clean[0]
firstcolumn = firstcolumn_str.strip()
firstcolumn_str_list.append(firstcolumn)
secondcolumn = flascard_clean[1]
flashcard_dict[firstcolumn] = secondcolumn

print
print ("###" * 3)

lista = list()
# this is version 4 - where lambda works but fails as it matches the string in all words.
# so if the word is "es"  all patterns are matched that has "es" AND NOT the specific word
filter_object = filter(lambda a: enter in a, firstcolumn_str_list)
for x in filter_object:
lista.append(x)

print (lista)
cc = 0 
if cc < len(lista):
for lambdatodictmatch in lista:
if lambdatodictmatch in flashcard_dict:
print (flashcard_dict[lambdatodictmatch])
else:
print ("NONEsense... nothing here")
else: 
print ("NONEsense... nothing here")

再次工作，但当我搜索汽车日产。我收到了四个回复，但我只收到了最后一个"粗糙"的回复。输出或我得到4个重复响应"粗糙"。

实现这一目标的最佳方法是什么?

Answer 1

如果你可能有重复的元素，那么你应该总是使用列表来保持单个值

if firstcolumn not in flashcard_dict: 
flashcard_dict[firstcolumn] = []
firstcolumn[firstcolumn].append(secondcolumn)

不是

flashcard_dict[firstcolumn] = secondcolumn

---

编辑:

包含其他更改的完整工作代码

• 首先，我使用了更短更易读的变量名
• 我在开始时读取文件，然后使用循环请求不同的卡。
• 我增加了!keys命令显示所有的键，!exit命令退出循环并完成程序，
• list(sorted(flashcards.keys()))给出字典中没有重复值(和排序)的所有键

我只使用io来模拟内存中的文件-所以每个人都可以简单地复制和运行这段代码(不创建文件FlashCards)，但你应该使用open(...)

text = '''car # automobile 4 wheels and run 
washington dc # the capital of United States 
fedora # an operating distro
cat file  # reads the file 
car nissan # altima
car nissan # altima ## first car 
car nissan # maxima
car nissan # rougue
'''
import io
# --- constansts --- 
DEBUG = True
# --- functions ---
def read_data(filename='FlashCards'):
if DEBUG:
print('[DEBUG] reading file')

flashcards = dict()  # with `s` at the end because it keeps many flashcards
#file_handler = open(filename)
file_handler = io.StringIO(text)

for line in file_handler:
line = line.strip()

parts = line.split("#", 1)

key   = parts[0].strip()
value = parts[1].strip()

if key not in flashcards:
flashcards[key] = []

flashcards[key].append(value)

all_keys = list(sorted(flashcards.keys()))
return flashcards, all_keys
# --- main ---
# - before loop -
# because words `key` and `keys` are very similar and it is easy to make mistake in code - so I added prefix `all_`
flashcards, all_keys = read_data()

print("#########")
# - loop -
while True:
print() # empty line to make output more readable
enter = input("Searching nemo (or command: !keys, !exit): ").strip().lower()
print() # empty line to make output more readable
if enter == '!exit':
break

elif enter == '!keys':
#print( "n".join(all_keys) )
for key in all_keys:
print('key>', key)

elif enter.startswith('!'):
print('unknown command:', enter)

else:
# keys which have `enter` only at 
#selected_keys = list(filter(lambda text: text.startswith(enter), all_keys))

# keys which have `enter` in any place (at the beginning, in the middle, at the end)
selected_keys = list(filter(lambda text: enter in text, all_keys))

print('selected_keys:', selected_keys)

if selected_keys:  # instead of `0 < len(selected_keys)`
for key in selected_keys:
# `selected_keys` has to exist in `flashcards` so there is no need to check if `key` exists in `flashcards`
print(key, '=>', flashcards[key])
else: 
print("NONEsense... nothing here")

# - after loop -
print('bye')