假设df1 a和b中的间隔不重叠，可以将pd.cut与pd.IntervalIndex一起使用：

import pandas as pd

# Your dfs here
df1 = pd.read_clipboard()
df2 = pd.read_clipboard()

idx = pd.IntervalIndex.from_arrays(df1["a"], df1["b"])
mapping = df1["c"].set_axis(idx)

df2["f"] = pd.cut(df2["d"], idx).map(mapping)

DF2：

d   e  f
0   3.3  10  1
1   5.5  20  1
2  14.5  11  3
3  17.2   5  5

您可以用途：

result = []
for item in df2['d']:
    for row in df1.iterrows():
        if row[1]['a'] <= item <= row[1]['b']:
            val = (row[1]['c'])
            break
        else:
            val = None
    result.append(val)
            
df2['f'] = result

print(df2)

import pandas as pd
df1 = pd.DataFrame({'a':[0,10,15],'b':[10,15,20],'c':[1,3,5]})
df2 = pd.DataFrame({'d':[3.3,5.5,9.5,17.2],'e':[10,20,11,5]})
df2['f']=0
for i in range(df2.shape[0]):
    for j in range(df1.shape[0]):
        if df2.d[i]>=df1.a[j] and df2.d[i]<=df1.b[j]:
            df2.f[i]=df1.c[j]
df2

这个选项怎么样？

# merge the two dfs
df = pd.merge(df2, df1, left_on='d', right_on='b', how='left')
df2['f'] = None
df2['f'] = df.apply(lambda x: x['c'] if x['a_x'] <= x['d'] <= x['b_x'] else None, axis=1)

如果没有重叠的间隔，则pd.IntervalIndexaccepted solution非常合适。
另一个选项是使用conditional_join from pyjanitor，它也可以处理重叠的间隔：

# pip install pyjanitor
import pandas as pd
import janitor
(df2
.conditional_join(
    # types have to be same
    # for columns to be compared
    df1.astype({"a":float, "b":float}), 
    ('d', 'a', '>='), 
    ('d', 'b','<='), 
    # depending on the data size,
    # numba may offer more performance
    use_numba=False,
    right_columns = {'c':'f'})
)
      d   e  f
0   3.3  10  1
1   5.5  20  1
2  14.5  11  3
3  17.2   5  5