同时需要从列表和元组值对的字典中删除项目 [英] Need to remove items from both a list and a dictionary of tuple value pairs at same time
问题描述
这与以前的问题,但我意识到我的目标要复杂得多:
我有一句话:福布斯亚洲200最佳下我有一个令牌像:
code> oldTokens = [u'Forbes',u'Asia',u'200',u'Best',u'Under',u'500',u'Billion',u'2011']
以前的解析器找出哪里应该有位置或号码位置的索引:
numberTokenIDs = {(7,):2011.0,(2)):200.0,(5,6):500000000000.00}
locationTokenIDs = {(0,1):u'Forbes Asia'}
令牌ID对应于令牌的位置或数字的索引,目的是获取一组新的令牌,如:
newTokens = [u'Asia',u'200',u'Best',u'Under',u'500',u'2011']
新的号码和位置tokenID可能是(为了避免索引超出范围的异常):
numberTokenIDs = {(5,):2011.0,(1)):200.0,(4):500000000000.00}
locationTokenIDs = {(0, ):u'Forbes Asia'}
本质上,我想通过一系列新的令牌,并能够最终创建一个新的句子:
LOCATION_SLOT NUMBER_SLOT NUMBER_SLOT NUMBER_SLOT以下
通过新的令牌,并用LOCATION_SLOT或NUMBER_SLOT替换正确的tokenID。如果我使用当前的一组号码和位置令牌ID,我将得到:
LOCATION_SLOT LOCATION_SLOT NUMBER_SLOT NUMBER_SLOT以下NUMBER_SLOT NUMBER_SLOT。
我该怎么做?
另一个例子是:
位置标记ID为:(0,1)
数字标记ID为:(3, 4)
旧样本标签[u'United',u'Kingdom',u'USD',u'1.240',u'billion']
我想要同时删除令牌,还可以更改位置和号码令牌ID,以便能够替换如下所示的句子:
sampleTokens [numberTokenID] =NUMBER_SLOT
sampleTokens [locationTokenID] =LOCATION_SLOT
这样替换的令牌是 [u'LOCATION_SLOT',u'USD',u'NUMBER_SLOT']
不是一个非常优雅,但工作的解决方案:
oldTokens = [u'Forbes',u'Asia',u'200',u'Best',u'Under',u'500',u'Billion',u'2011']
numberTokenIDs = {(7,):2011.0,(2)):200.0,(5 ,6):500000000000.00}
locationTokenIDs = {(0,1):u'Forbes Asia'}
newTokens = []
newnumberTokenIDs = {}
newlocationTokenIDs = {}
new_ind = 0
skip = False
for range(len(oldTokens)):
如果跳过:
skip = False
继续
for loc_ind在locationTokenIDs.keys()中:
if ind in loc_ind:
newTokens.append(oldTokens [ind + 1])
newlocationTokenIDs [(new_ind,]] = locationTokenIDs [loc_ind]
new_ind + = 1
如果len(loc_ind)> 1:#如果元组中有2个元素,则跳过下一个位置
skip = True
break
else:
在numberTokenIDs.keys()中的num_ind:
if ind in num_ind:
newTokens.append(oldTokens [ind])
newnumberTokenIDs [(new_ind,]] = numberTokenIDs [num_ind]
new_ind + = 1
如果len(num_ind) > 1:
skip = True
break
else:
newTokens.append(oldTokens [ind])
new_ind + = 1
newTokens
Out [37]:[u'Asia',u'200',u'Best',u'Under',u'500',u'2011']
newnumberTokenIDs
输出[38]:{(1,):200.0,(4,):500000000000.0,(5,):2011.0}
newlocationTokenIDs
Out [39] 0,):u'Forbes Asia'}
This is very related to a previous question but I realised that my objective is much more complicated:
I have a sentence: "Forbes Asia 200 Best Under 500 Billion 2011"
I have tokens like:
oldTokens = [u'Forbes', u'Asia', u'200', u'Best', u'Under', u'500', u'Billion', u'2011']
And the indices of where a previous parser has figured out where there should be location or number slots:
numberTokenIDs = {(7,): 2011.0, (2,): 200.0, (5,6): 500000000000.00}
locationTokenIDs = {(0, 1): u'Forbes Asia'}
The token IDs correspond to the index of the tokens where there are locations or numbers, the objective is to obtain a new set of tokens like:
newTokens = [u'Asia', u'200', u'Best', u'Under', u'500', u'2011']
With new number and location tokenIDs perhaps like (to avoid index out of bounds exceptions):
numberTokenIDs = {(5,): 2011.0, (1,): 200.0, (4,): 500000000000.00}
locationTokenIDs = {(0,): u'Forbes Asia'}
Essentially I would like to go through the new reduced set of tokens, and be able to ultimately create a new sentence called:
"LOCATION_SLOT NUMBER_SLOT Best Under NUMBER_SLOT NUMBER_SLOT"
via going through the new set of tokens and replacing the correct tokenID with either "LOCATION_SLOT" or "NUMBER_SLOT". If I did this with the current set of number and location token IDs, I would get:
"LOCATION_SLOT LOCATION_SLOT NUMBER_SLOT Best Under NUMBER_SLOT NUMBER_SLOT NUMBER_SLOT".
How would I do this?
Another example is:
Location token IDs are: (0, 1)
Number token IDs are: (3, 4)
Old sampleTokens [u'United', u'Kingdom', u'USD', u'1.240', u'billion']
Where I want to both delete tokens and also change location and number token IDs to be able to replace the sentence like:
sampleTokens[numberTokenID] = "NUMBER_SLOT"
sampleTokens[locationTokenID] = "LOCATION_SLOT"
Such that the replaced tokens are [u'LOCATION_SLOT', u'USD', u'NUMBER_SLOT']
Not a very elegant, but working solution:
oldTokens = [u'Forbes', u'Asia', u'200', u'Best', u'Under', u'500', u'Billion', u'2011']
numberTokenIDs = {(7,): 2011.0, (2,): 200.0, (5,6): 500000000000.00}
locationTokenIDs = {(0, 1): u'Forbes Asia'}
newTokens = []
newnumberTokenIDs = {}
newlocationTokenIDs = {}
new_ind = 0
skip = False
for ind in range(len(oldTokens)):
if skip:
skip=False
continue
for loc_ind in locationTokenIDs.keys():
if ind in loc_ind:
newTokens.append(oldTokens[ind+1])
newlocationTokenIDs[(new_ind,)] = locationTokenIDs[loc_ind]
new_ind += 1
if len(loc_ind) > 1: # Skip next position if there are 2 elements in a tuple
skip = True
break
else:
for num_ind in numberTokenIDs.keys():
if ind in num_ind:
newTokens.append(oldTokens[ind])
newnumberTokenIDs[(new_ind,)] = numberTokenIDs[num_ind]
new_ind += 1
if len(num_ind) > 1:
skip = True
break
else:
newTokens.append(oldTokens[ind])
new_ind += 1
newTokens
Out[37]: [u'Asia', u'200', u'Best', u'Under', u'500', u'2011']
newnumberTokenIDs
Out[38]: {(1,): 200.0, (4,): 500000000000.0, (5,): 2011.0}
newlocationTokenIDs
Out[39]: {(0,): u'Forbes Asia'}
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