好的，再次感谢@Laurent，他指出了我忽略的一个组合，所以我意识到我需要遍历所有可能的潜在值组合，而不是试图比较过滤器是最小值还是最大值。

import pandas as pd
import itertools
import numpy as np

### On two variables
df = pd.DataFrame({'var_1_lower_limit':[1,2,1,3,2],'var_1_upper_limit':[5,5,4,7,6],'var_2_lower_limit':[5,5,3,6,4],'var_2_upper_limit':[9,9,8,9,7]})
df.sort_values(['var_1_lower_limit','var_1_upper_limit','var_2_lower_limit','var_2_upper_limit'],ascending=True, inplace=True) # I think important to sort here
display(df)

default_possibilities = [0,1,2,3,4,5,6,7,8,9]
possibilities = list(itertools.product(default_possibilities,default_possibilities,default_possibilities,default_possibilities)) #Creates all combinations of possibilities 0-9,0-9,0-9,0-9
print('length of possibilities',len(possibilities))
possibilities = [item for item in possibilities if item[1]>=item[0] and item[3]>=item[2]] # removes combinations that don't make sense such as first value is 5 and second value is 3
print('length of revised possibilities',len(possibilities))

filtering_output = []

# For each combination of potential values, see if it passes a filter
for i in possibilities: 
  a = i[0]
  b = i[1]
  c = i[2]
  d = i[3]
  for z in df.itertuples():
    if z.var_1_lower_limit <=a and z.var_1_upper_limit >=b and z.var_2_lower_limit <=c and z.var_2_upper_limit >=d:
      filtering_output.append({'combination':i,'filter_id':str(z.var_1_lower_limit)+','+str(z.var_1_upper_limit)+','+str(z.var_2_lower_limit)+','+str(z.var_2_upper_limit),'var_1_lower_limit':z.var_1_lower_limit,'var_1_upper_limit':z.var_1_upper_limit, 'var_2_lower_limit':z.var_2_lower_limit,'var_2_upper_limit':z.var_2_upper_limit,'pass_or_fail':'YES'})
    else:
      continue

filtering_output_df = pd.DataFrame(filtering_output) ### This df has all passing combinations
display(filtering_output_df)  

unique_combinations = filtering_output_df.drop_duplicates(subset=['combination'], keep='first') # Since there are combinations that pass multiple filters, reduce to only the first iteration of that combination

final_filters = unique_combinations.drop_duplicates(subset=['var_1_lower_limit','var_1_upper_limit','var_2_lower_limit','var_2_upper_limit'], keep='first') # since df was sorted, keeping only filters that had a passing combination
display(final_filters)

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下面是一种使用Pandas shift的惯用方法：

import pandas as pd

# Sort dataframe and set 'unique' column with 'True' as default value
df = df.sort_values(
    by=[
        "var_1_lower_limit",
        "var_1_upper_limit",
        "var_2_lower_limit",
        "var_2_upper_limit",
    ]
).assign(unique=True)

# Identify filters not contained within others by comparing current row
# with previous one or the next, and set them as 'False'
non_unique_filters = (
    (df["var_1_lower_limit"] >= df["var_1_lower_limit"].shift(-1))
    & (df["var_1_upper_limit"] <= df["var_1_upper_limit"].shift(-1))
    & (df["var_2_lower_limit"] >= df["var_2_lower_limit"].shift(-1))
    & (df["var_2_upper_limit"] <= df["var_2_upper_limit"].shift(-1))
) | (
    (df["var_1_lower_limit"] >= df["var_1_lower_limit"].shift(1))
    & (df["var_1_upper_limit"] <= df["var_1_upper_limit"].shift(1))
    & (df["var_2_lower_limit"] >= df["var_2_lower_limit"].shift(1))
    & (df["var_2_upper_limit"] <= df["var_2_upper_limit"].shift(1))
)
df.loc[non_unique_filters, "unique"] = False

df = df.sort_index()

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然后又道：

print(df)
# Output
   var_1_lower_limit  var_1_upper_limit  var_2_lower_limit  var_2_upper_limit  unique
0                  1                  5                  5                  9    True
1                  2                  5                  5                  9   False
2                  1                  4                  3                  8    True
3                  3                  7                  6                  9    True
4                  2                  6                  4                  7    True

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