Question

有没有办法像str.contains一样使用字典？

我有一本字典，它看起来像这样：

A = {'Hey':1}

我知道对于字典，它会查找完全匹配（区分大小写和空格），因此我不知道是否可以执行此操作。

我的数据框看起来像这样：

          Statements
0    Hey how are you?
1    Hey is their anyway to find that
2    Hey is their a way to prove this
3    over their, hey, how are you?

我想要做的是使用我的字典并基本查看语句中的每一行，如果字符串包含Hey将其更改为1，此外，如果我可以这样做，我想知道是否可以在其中放置多个语句字典？像这样：

A = {＆＃39;嘿＆＃39;，＆＃39;你好＆＃39;，＆＃39;嗨＆＃39; ：1}

我想要做的是在字典中放入一堆可能的字符串，如果在语句中找到这些字符串，则根据需要进行更改。在这个例子中，Hey是语句中唯一可以改变的单词。

我的预期结果如下：

          Statements
0    1 how are you?
1    1 is their anyway to find that
2    1 is their a way to prove this
3    over their, 1, how are you?

Answer 1

我认为您可以先创建列表词典，然后将keys与values交换为d和replace：

L = ['Hey how are you?',
     'Hey is their anyway to find that',
     'Hey is their a way to prove this',
     'over their, hey, how are you?']

df = pd.DataFrame({'Statements':L})

A = {'1':['Hey', 'Hello', 'Hi', 'hey']}
d = {k: oldk for oldk, oldv in A.items() for k in oldv}
print (d)
{'Hi': '1', 'hey': '1', 'Hey': '1', 'Hello': '1'}

df['Statements'] = df['Statements'].replace(d, regex=True)
print (df)
                       Statements
0                  1 how are you?
1  1 is their anyway to find that
2  1 is their a way to prove this
3     over their, 1, how are you?

Answer 2

有更好的方法可以做到这一点，但想在进入会议之前发帖:)

首先，逻辑将如下工作 1）循环遍历列表中的每个元素 2）拆分列表中的句子 3）循环句子中的每个单词并检查数学 4）如果匹配则更新索引

假设a是字典

for line in a:
    sentence = line.split()
    for word in sentence:
        if word == 'Hey':
           # Found an index which contains the word so update the index here
           a[line] = 'New value for index'

Answer 3

执行此操作的一点点方法，但允许您在一行中检查您想要的任意数量的单词。

for i in range(df.shape[0]):
    line_split = df['Statements'][i].split()
    for j in range(len(line_split)):
        if line_split[j] in (['Hey', 'hey']):
           line_split[j] = '1'
    df['Statements'][i] = ' '.join(line_split)

使用词典在一个语句Python中一次更改多个关键字

3 个答案: