考虑到被比较的行可能会根据每次比较的结果而改变，我不确定你是否可以在不使用一些等同于for循环的逻辑的情况下实现这一点：

#Taking initial comparison values from first row
b,c = df.iloc[0][['b','c']]
#Including first row in result
filters = [True]

#Skipping first row in comparisons
for index, row in df.iloc[1:].iterrows():
    if 0.99*b <= row['b'] <= 1.01*b and 0.99*c <= row['c'] <= 1.01*c:
        filters.append(False)
    else:
        filters.append(True)
        # Updating values to compare based on latest accepted row
        b = row['b']
        c = row['c']

df2 = df.loc[filters]

print(df2)

   a      b       c
0  1  20.02   10.12
2  3  19.98  123.54
4  5  26.75  245.12
5  6  56.12  895.21

检查边缘情况，其中行（n+1）接近行（n）（并且被排除），但是行（n+2）接近行（n+1）但不接近行（n）（并且因此应该被包括）：

df = pd.DataFrame({'a':[1,2,3], \
                   'b':[20,20,20],\
                   'c':[100,100.9,101.1]})

a   b      c
0  1  20  100.0
2  3  20  101.1

很大程度上基于ukemi之前的回答。在这个例子中，每个列的值都与之前接受的所有行进行比较，而不仅仅是最后接受的行。

df = pd.DataFrame({'a':[1,2,3,4,5,6,7,8,9],'b':[20.02,19.96,19.98,20.10,26.75,56.12, 20.04,56.24, 56.15],\
               'c':[10.12,10.10,123.54,124.12,245.12,6.00,10.11,6.50,128.67]})

    a     b       c
0   1   20.02   10.12
1   2   19.96   10.10
2   3   19.98   123.54
3   4   20.10   124.12
4   5   26.75   245.12
5   6   56.12   6.00
6   7   20.04   10.11
7   8   56.24   6.50
8   9   56.15   128.67

b = []
c = []

#Taking initial comparison values from first row
b.append(df.iloc[0]['b'])
c.append(df.iloc[0]['c'])

#Including first row in result
filters = [True]

#Skipping first row in comparisons
for index, row in df.iloc[1:].iterrows():
    tag = 0
    for i in range(len(b)):
        #Thresholds have been changed to 5% and 10% respectively in this case.
        if 0.95*b[i] <= row['b'] <= 1.05*b[i] and 0.90*c[i] <= row['c'] <= 1.10*c[i]:
            filters.append(False)
            tag = 1
            break

    if tag == 0:
        filters.append(True)
        # Updating values to compare based on latest accepted row
        b.append(row['b'])
        c.append(row['c'])

df2 = df.loc[filters]

print(df2)

    a    b       c
0   1   20.02   10.12
2   3   19.98   123.54
4   5   26.75   245.12
5   6   56.12   6.00
8   9   56.15   128.67

请告诉我是否有更快的方法来达到同样的效果。

我使用itertools.pairwise，它在python〉=3.10中可用。
1.对行进行排序;
1.按相反顺序成对比较;
1.如果在标准内，则丢弃较高的一个。
假设列名为key，删除条件为EPISION。

from itertools import pairwise
from tqdm import tqdm

df = ...
key = "key"
EPISION = 1e-4

df = df.sort_values(by=[key])
for (i, iser), (j, jser) in tqdm(df[::-1].iterrows(), total=len(df) - 1):
    if iser[key] - jser[key] < EPISION:
        df.drop(index=i, inplace=True)

print(df)