Python 制作缩写-选择非停止词的第一个字符

给出一个停止字列表和一个数据框，其中1列具有完整的格式，如图所示-

stopwords = ['of', 'and', '&', 'com', 'org']
df = pd.DataFrame({'Full form': ['World health organization', 'Intellectual property', 'royal bank of canada']})
df

+---+---------------------------+
|   |         Full form         |
+---+---------------------------+
| 0 | World health organization |
| 1 | Intellectual property     |
| 2 | Royal bank of canada      |
+---+---------------------------+

我正在寻找一种方法，使相邻列的缩写忽略stopwords（如果有的话）

预期产出：

+---+---------------------------+----------------+
|   |         Full form         |   Abbreviation |
+---+---------------------------+----------------+
| 0 | World health organization |   WHO          |
| 1 | Intellectual property     |   IP           |
| 2 | Royal bank of canada      |   RBC          |
+---+---------------------------+----------------+

这应该做到：

import pandas as pd

stopwords = ['of', 'and', '&', 'com', 'org']
df = pd.DataFrame({'Full form': ['World health organization', 'Intellectual property', 'royal bank of canada']})


def abbrev(t, stopwords=stopwords):
    return ''.join(u[0] for u in t.split() if u not in stopwords).upper()


df['Abbreviation'] = df['Full form'].apply(abbrev)

print(df)

输出

                   Full form Abbreviation
0  World health organization          WHO
1      Intellectual property           IP
2       royal bank of canada          RBC

这应该做到：

import pandas as pd

stopwords = ['of', 'and', '&', 'com', 'org']
df = pd.DataFrame({'Full form': ['World health organization', 'Intellectual property', 'royal bank of canada']})


def abbrev(t, stopwords=stopwords):
    return ''.join(u[0] for u in t.split() if u not in stopwords).upper()


df['Abbreviation'] = df['Full form'].apply(abbrev)

print(df)

输出

                   Full form Abbreviation
0  World health organization          WHO
1      Intellectual property           IP
2       royal bank of canada          RBC

另一种方法：

df['Abbreviation'] = (df['Full form'].replace(stopwords, '', regex=True)
                      .str.split()
                      .apply(lambda word: [l[0].upper() for l in word])
                      .str.join(''))

另一种方法：

df['Abbreviation'] = (df['Full form'].replace(stopwords, '', regex=True)
                      .str.split()
                      .apply(lambda word: [l[0].upper() for l in word])
                      .str.join(''))

---

下面是一个正则表达式解决方案：

stopwods = ['of', 'and', '&', 'com', 'org']
stopwords_re = r"(?!" + r"\b|".join(stopwords) + r"\b)"
abbv_re = r"\b{}\w".format(stopwords_re)

def abbrv(s):
    return "".join(re.findall(abbv_re, s)).upper()

[out]：

>>> abbrv('royal bank of scotland')
'RBS'

与熊猫一起使用：

df['Abbreviation'] = df['Full form'].apply(abbrv)

---

有关正则表达式的完整解释，请参见：

简言之

• \b{}\w

  ：查找单词边界后的所有字符

• （？！of\b| and\b|&\b）

  ：除非它在停止词列表中

---

这里有一个正则表达式解决方案：

stopwods = ['of', 'and', '&', 'com', 'org']
stopwords_re = r"(?!" + r"\b|".join(stopwords) + r"\b)"
abbv_re = r"\b{}\w".format(stopwords_re)

def abbrv(s):
    return "".join(re.findall(abbv_re, s)).upper()

[out]：

>>> abbrv('royal bank of scotland')
'RBS'

与熊猫一起使用：

df['Abbreviation'] = df['Full form'].apply(abbrv)

---

有关正则表达式的完整解释，请参见：
简言之

• \b{}\w

  ：查找单词边界后的所有字符

• （？！of\b| and\b|&\b）

  ：除非它在停止词列表中

---

太快了。谢谢@Danielt这很快。谢谢@Daniel