如果值在此列表中并且在列值的某个位置，我可以用列表值替换该列值吗？熊猫数据框

Question

有很多关于这方面的文档，但我无法弄清楚。

这是一个列表，我需要检查这些值之一是否在我的列值中。如果是这样，用列表值替换整个单元格。

active_crews = ["CREW #101", "CREW #102", "CREW #203", "CREW #301", "CREW #404", "CREW #501", "CREW #406", "CREW #304", "CREW #701", "CREW #702", "CREW #703", "CREW #704", "CREW #705", "CREW #706",
                "CREW #707" "CREW #708", "CREW #801", "CREW #802", "CREW #803", "CREW #805"]

我要替换的数据示例。是的，格式也有细微的差别：

Debris Crew WO# 
REFER TO IAP 12/16 TO 12/19 CREW #405
REFER TO IAP 06/02 TO 06/05 CREW #406
REFER TO IAP 03/24TO 03/27 CREW # 803

预期输出

Debris Crew WO#
CREW #405
CREW #406
CREW #803

我的问题是我不知道如何告诉 python 使用列表搜索列值以查找匹配项。并且该列表值是否在该列值中。用列表值替换当前列值

我尝试过的代码：

1)

df.loc[df['Debris Crew WO#'] == active_crews, 'Debris Crew WO#']
# doesn't work. This was done before research lol I get the following error, which makes sense
# ValueError: ('Lengths must match to compare', (2216,), (19,))

df.loc[:, ['Place Holder']] = df.loc[:, 'Debris Crew WO#'].str[28:]
# this code "works" but due to different formatting i get data back like this:
8   REW #406
9   CREW #406
# not very effective and can not be relied on. I hate hard coding anything.

df.loc[:, ['Place Holder']] = df.loc[:, 'Debris Crew WO#'].str[26:]
df.loc[:, ['Place Holder']] = df[['Place Holder']].str.split().join(" ")
# tried this due to I have this filter for specials characters with a for loop in a different code and yet I get this error and I have no clue why. Works on my other codes with no problems

#AttributeError: 'DataFrame' object has no attribute 'str'

# even if I use .loc I get the same error:
df.loc[:, ['Place Holder']] = df.loc[:, 'Debris Crew WO#'].str[26:]
df.loc[:, ['Place Holder']] = df.loc[:, ['Place Holder']].str.split().join(" ")
#plus its still hard coding (gross)

接下来我将与 RE 合作。有人告诉我，它非常适合像过滤类型这样的“CTRL 查找”样式，并且是数据科学中的关键工具。因此，在下周+从 RE Documentation 开始，并在这个问题上练习它。随着我的进展，将编辑更新

就是这么说的。我已经学习python将近两个月了。请原谅任何只是尝试和试验的“菜鸟”风格/编码，这样我就可以让我的生活和我周围的其他人变得更好。 任何帮助将不胜感激。提前致谢

Answer 1

引用列表的方法#1：

您可以将str.extract() 与作为join('|') 加入列表的捕获组一起使用。 | 符号用于 OR，允许您同时搜索每一行的多个值。捕获组需要在它们周围加上括号，这就是我在前后添加括号作为字符串的原因。

active_crews = ["CREW #101", "CREW #102", "CREW #203", "CREW #301", "CREW #404", "CREW #501", 
                "CREW #406", "CREW #304", "CREW #701", "CREW #702", "CREW #703", "CREW #704", 
                "CREW #705", "CREW #706", "CREW #707" "CREW #708", "CREW #801", "CREW #802", 
                "CREW #803", "CREW #805"]

df['Debris Crew WO#'] = df['Debris Crew WO#'].str.extract('(' + '|'.join(active_crews) + ')')
df

#You  can also use a formatted string like this:
df['Debris Crew WO#'] = df['Debris Crew WO#'].str.extract(f'({"|".join(active_crews)})')

Out[1]: 
  Debris Crew WO#
0             NaN
1       CREW #406
2             NaN

---

方法 #2 基于正则表达式模式提取并忽略列表。空格后的? 表示该空格是可选的。除了space，您还可以对多个空格执行\s 或\s+。 \d+ 表示连续的数字。如果数字中有逗号，则正则表达式略有不同：

df['Debris Crew WO#'] = df['Debris Crew WO#'].str.extract('(CREW ?# ?\d+)')

Out[2]: 
  Debris Crew WO#
0            #405
1            #406
2           # 803