用途：

phrase = 'liability commercial'

#match by substrings - splitted values by spaces
m = df['Text'].str.contains(phrase.replace(' ','|'))
#match by splitted values by spaces
m = df['Text'].isin(phrase.split())

#filter rows by mask and get last duplicated values in Text column
df = df[m].drop_duplicates(['Text'], keep='last')
print (df)
   id  left  top  width  height        Text
2   3     1   47      9       4   liability
3   4    10   69     32      67  commercial

字符串
或者，如果需要按条件更改掩码按匹配行分组，则此处拆分值的位置和可能的重复不计数：

phrase = 'liability commercial'
m = ~df['Text'].str.contains(phrase.replace(' ','|'))
#m = ~df['Text'].isin(phrase.split())

df = df[m.cumsum().duplicated(keep=False) & ~m]
print (df)
   id  left  top  width  height        Text
2   3     1   47      9       4   liability
3   4    10   69     32      67  commercial

型
如果需要通过拆分值进行精确匹配，则可以修改this solution：

phrase = 'liability commercial'

#https://stackoverflow.com/a/49005205/2901002
pat = np.asarray(phrase.split())
N = len(pat)

def rolling_window(a, window):
    shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
    strides = a.strides + (a.strides[-1],)
    c = np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
    return c

arr = df['Text'].values
b = np.all(rolling_window(arr, N) == pat, axis=1)
c = np.mgrid[0:len(b)][b]

d = [i  for x in c for i in range(x, x+N)]
df = df[np.in1d(np.arange(len(arr)), d)]
print (df)
   id  left  top  width  height        Text
2   3     1   47      9       4   liability
3   4    10   69     32      67  commercial

型