你可以numpy广播：

# Convert to numpy
vals1 = df_test_1.values
vals2 = df_test_2.values

# Remove from df_test_1
arr1 = np.abs(vals1 - vals2[:, None])
msk1 = ~np.any(np.all(arr1 < [100, 100, 0.1], axis=2), axis=1)

# Remove from df_test_2
arr2 = np.abs(vals2 - vals1[:, None])
msk2 = ~np.any(np.all(arr1 < [100, 100, 0.1], axis=2), axis=1)

out = pd.concat([df_test_1[msk1], df_test_2[msk2]], ignore_index=True)

输出：

>>> out
       x      y       z
0  151.0  242.0  124.42
1  633.0  176.0  875.76

评论@James
这将删除left vs right和right vs left，但不会删除left vs left或right vs right中的重复项。
在这种情况下：

df_test_3 = pd.concat([df_test_1, df_test_2])

arr = df_test_3.values
msk = np.abs(arr - arr[:, None]) < [100, 100, 0.1]
out = df_test_3[np.sum(np.all(msk, axis=2), axis=1) == 1]
print(out)

# Output
       x      y       z
3  151.0  242.0  124.42
3  633.0  176.0  875.76

要在阈值内对数据框进行重复数据消除，您需要计算每列中每个值之间的差异，并查看这些值是否在阈值差异范围内。这是针对任何 Dataframe 的通用解决方案。

from itertools import combinations

df = df_test_3.reset_index(drop=True)

# using combinations ensures a lower-triangular matrix of comparison indices
mi = MultiIndex.from_tuples(combinations(df.index, 2))
ix_left = mi.get_level_values(0)
ix_right = mi.get_level_values(1)
df_cross_diff = df.loc[ix_left].set_index(mi) - df.loc[ix_right].set_index(mi)

# that mask is comparisons that are considered duplicates.
drop_mask = pd.concat([
    df_cross_diff.x.abs().le(100),
    df_cross_diff.y.abs().le(100),
    df_cross_diff.z.abs().le(0.1),
], axis=1).all(axis=1)

# this extracts the pairs of indices that are considered duplicates:
dupes_mi = drop_mask.loc[drop_mask].index
dupes_left = matches_mi.get_level_values(0)
dupes_right = matches_mi.get_level_values(1)

# and finally we remove the duplicates.
df_deduped = df.drop(dupes_right)