我正在寻找一种以排序的方式对所有变量及其值进行腌制的方法。我正在尝试以下方法:
预期结果:将va
的值附加到totalList
,然后追加vb's
的值,依此类推。然后将totalList
转储到泡菜文件中。
import pickle
import time
import os
import re
v0 = ["r0", "hello","hello there","hi there","hi", "hey","good to see you"]
v1 = ["r1", "how are you", "how do you do","how are you doing", "how you doing"]
v2 = ["r2", "What's up" , "what is up" , "what's up bro"]
v10 = ['r10', 'who made you', 'who built you', 'who is your creator']
imports = "pickle","time","os","re"
totalList = []
for key in dir():
if key.startswith("__") == 0 and key not in imports:
print(globals()[key])
totalList.append(globals()[key])
# print(totalList) # getting weird values
with open('queryList.pkl', 'wb') as f:
pickle.dump(totalList, f)
我得到这样的结果:
>> ('pickle', 'time', 'os', 're')
>> []
>> ['r0', 'hello', 'hello there', 'hi there', 'hi', 'hey', 'good to see you']
>> ['r1', 'how are you', 'how do you do', 'how are you doing', 'how you doing']
>> ['r10', 'who made you', 'who built you', 'who is your creator']
>> ['r2', "What's up", 'what is up', "what's up bro"]
我想摆脱结果('pickle', 'time', 'os', 're')
和[]
并在追加到totalList
之前对结果进行排序,或者在迭代之前对其进行排序。
答案 0 :(得分:1)
这是一种执行所需操作的方法。您应该忽略一些您不感兴趣的变量-特别是imports
和totalList
。
ignore_list = [ "ignore_list", "totalList"]
totalList = []
for key in dir():
if not key.startswith("__") and key not in ignore_list and type(globals()[key]) != type(pickle):
print(key)
totalList.append(globals()[key])
totalList = sorted(totalList, key=lambda x: int(x[0][1:]) )
结果是:
[['r0', 'hello', 'hello there', 'hi there', 'hi', 'hey', 'good to see you'],
['r1', 'how are you', 'how do you do', 'how are you doing', 'how you doing'],
['r2', "What's up", 'what is up', "what's up bro"],
['r10', 'who made you', 'who built you', 'who is your creator']]