您可以编写一个函数来检查DF值是否包含任何字符串，如果包含，则更改colors列。注意下面的代码丢失了a' pink中的空格e以生成所需的结果。我不确定这是否更广泛地适用于您的语言结构。

import pandas as pd

color_list = ['red', 'black', 'white', "a'pink", 'blue']
pattern = '|'.join(color_list)

data = ['on red', 'aup red gone', 'black with bi', "ao a' pink", 'dgfh blu', 'a black pen']

df = pd.DataFrame(data, columns=['colors'])

def func(row):
    for word in color_list:
    if word in row['colors'].replace("' ", "'"):
            row['is it'] = True
            row['colors'] = word
            break
    else:
        row['is it'] = False
    return row

df = df.apply(func, axis = 1)

print(df)

其给出：

colors  is it
0       red   True
1       red   True
2     black   True
3    a'pink   True
4  dgfh blu  False
5     black   True

您应该能够合并字典D={ }。
为便于说明：

>>>
>>> import re
>>>
>>> D = {}
>>> D['red'] = 'True'
>>> D['black'] = 'True'
>>> D["a' pink"] = 'True'
>>> D['dgfh blu'] = 'False'
>>>
>>> column = '''
...          on red
...    aup red gone
...   black with bi
...      ao a' pink
...        dgfh blu
...     a black pen
... '''
>>>
>>> def repl(m):
...   return m.group(1)+"  "+D[m.group(1)]
...
>>> column = re.sub(r"(?m)^.*?(red|black|a'[ ]pink|dgfh[ ]blu)\b.*?$",repl,column)
>>>
>>> print(column)

red  True
red  True
black  True
a' pink  True
dgfh blu  False
black  True

>>>