修改代码以处理一长串字符串

Question

我想为一项任务准备一长串数据。 我已经能够将在单个实例上完成任务的代码放在一起，但现在我想让它通过一个列表运行。以下是我尝试过的。

用于测试的单个实例.....

sentences = ['if the stimulus bill had become hamstrung by a filibuster threat or recalcitrant conservadems']
antecedents = ['bill had become hamstrung by']

实际用例是 pandas 数据框中的两列，我已将其转换为列表

f = tra_df['sentence'].tolist()
b = tra_df['antecedent'].tolist()

单个用例的代码....

results =[]

ous = 1
ayx = ' '.join([str(elem) for elem in antecedents])
ayxx = ayx.split(" ")
antlabels = []    
for i in range(len(ayxx)):

    antlabels.append(ous)
    lab = ' '.join([str(elem) for elem in antlabels])



     # Build the regex string required
rx = '({})'.format('|'.join(re.escape(el) for el in antecedents))
     # Generator to yield replaced sentences
it = (re.sub(rx, lab, sentence) for sentence in sentences)
     # Build list of paired new sentences and old to filter out where not the same
results = ([new_sentence for old_sentence, new_sentence in zip(sentences, it) if old_sentence != new_sentence])

# replace other non 1 values with 0
nw_results = ' '.join([str(elem) for elem in results])
ew_results= nw_results.split(" ")
new_results = ['0' if i is not '1' else i for i in ew_results]
labels =([int(e) for e in new_results]) 

labels

这就是我得到的结果

[0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0]

对大列表稍作修改的代码

for sentences, antecedents in zip(f, b):
    gobels = []
    #def format_labels(antecedents,sentences):
    results =[]
    #lab =[]
    ous = 1
    ayx = ' '.join([str(elem) for elem in antecedents])
    ayxx = ayx.split(" ")
    antlabels = []    
    for i in range(len(ayxx)):
        antlabels.append(ous)
        lab = ' '.join([str(elem) for elem in antlabels])



     # Build the regex string required
    rx = '({})'.format('|'.join(re.escape(el)for el in antecedents))
     # Generator to yield replaced sentences
    it = (re.sub(rx, lab, sentence)for sentence in sentences)
     # Build list of paired new sentences and old to filter out where not the same
    results = ([new_sentence for old_sentence, new_sentence in zip(sentences, it) if old_sentence != new_sentence])

    nw_results = ' '.join([str(elem) for elem in results])
    ew_results= nw_results.split(" ")
    new_results = ['0' if i is not '1' else i for i in ew_results]
    labels =([int(e) for e in new_results]) 

    t2 = time.time()
    gobels.append(labels)

现在，我得到的不是包含 0 和 1 的字符串列表，而是只有 1 的长列表.....

可能出了什么问题？

[[1,
  1,
  1,
  1,
  1,
  1,
  1,
  1,
  1,
 ........]

Answer 1

这样的东西可能会更好地扩展。可能还有一种更 Pythonic 的方式来做这件事。

a = '1 2 3 4 5'
b = '3 4 6'

a = a.split()
b = b.split()

for idx, val in enumerate(b):
    try:
        a[a.index(val)] = True
    except ValueError:
        pass

for idx, val in enumerate(a):
    if val is not True:
        a[idx] = False

print([1.0 if i else 0.0 for i in a])
# [0.0, 0.0, 1.0, 1.0, 0.0]