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 = 1ayx = ' '.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 requiredrx = '({})'.format('|'.join(re.escape(el) for el in antecedents))     # Generator to yield replaced sentencesit = (re.sub(rx, lab, sentence) for sentence in sentences)     # Build list of paired new sentences and old to filter out where not the sameresults = ([new_sentence for old_sentence, new_sentence in zip(sentences, it) if old_sentence != new_sentence])# replace other non 1 values with 0nw_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]大列表的稍微修改的代码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])现在，我得到一个只有 1 的长列表，而不是包含 0 和 1 的字符串列表......有什么问题？

Answer 1

像这样的东西可能会更好地扩展。可能还有一种更 Pythonic 的方式来执行此操作。

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]