英文:
Pandas isin() not working properly with numerical values
问题
我有一个pandas数据帧，其中一列全是浮点数，另一列包含浮点数列表、None或只是浮点数值。我已确保所有值都是浮点数。
最终，我想使用pd.isin()来检查value_2中有多少记录的value_1，但对我来说没有效果。当我运行下面的代码时：
df[~df['value_1'].isin(df['value_2'])]
下面是它返回的结果，这不是预期的，因为显然value_1中的某些值在value_2列表中：
value_1 value_20 88870.0 [88870.0]1. 150700.0 None2 225000.0 [225000.0, 225000.0]3. 305000.0 [305606.0, 305000.0, 1067.5]4 392000.0 [392000.0]5 198400.0 396
我漏掉了什么？请帮助。
英文:
I have a pandas dataframe where one column is all float, another column either contains list of floats, None, or just float values. I have ensured all values are floats.
Ultimately, I want to use pd.isin() to check how many records of value_1 are in value_2 but it is not working for me. When I ran this code below:
df[~df['value_1'].isin(df['value_2'])]
This below is what it returned which is not expected since clearly some values in value_1 are in the value_2 lists.:
0 88870.0 [88870.0]1. 150700.0 None2 225000.0 [225000.0, 225000.0]3. 305000.0 [305606.0, 305000.0, 1067.5]4 392000.0 [392000.0]5 198400.0 396
What am I missing? Please help.
答案1
得分: 2
您可以在列表推导式中使用numpy.isin进行布尔索引：
import numpy as npout = df[[bool(np.isin(v1, v2)) for v1, v2 in zip(df['value_1'], df['value_2'])]]
输出：
value_1 value_20 88870.0 [88870.0]2 225000.0 [225000.0, 225000.0]3 305000.0 [305606.0, 305000.0, 1067.5]4 392000.0 [392000.0]
英文:
You can use boolean indexing with numpy.isin in a list comprehension:
import numpy as npout = df[[bool(np.isin(v1, v2)) for v1, v2 in zip(df['value_1'], df['value_2'])]]
Output:
value_1 value_20 88870.0 [88870.0]2 225000.0 [225000.0, 225000.0]3 305000.0 [305606.0, 305000.0, 1067.5]4 392000.0 [392000.0]
答案2
得分: 1
使用zip和列表推导来测试列表是否不包含浮点数，如果不包含浮点数，则通过传递False来删除行，使用布尔索引进行筛选：
df = pd.DataFrame({'value_1':[88870.0,150700.0,392000.0],'value_2':[[88870.0],None, [88870.0,45.4]]})print(df)
输出：
value_1 value_20 88870.0 [88870.0]1 150700.0 None2 392000.0 [88870.0, 45.4]
针对测试标量值的需求：
mask = [a not in b if isinstance(b, list) else a != bfor a, b in zip(df['value_1'], df['value_2'])]df2 = df[mask]print(df2)
输出：
value_1 value_21 150700.0 None2 392000.0 [88870.0, 45.4]
性能方面：纯Python应该更快，最好在真实数据中进行测试：
# 20k行N = 10000df = pd.DataFrame({'value_1':[88870.0,150700.0,392000.0] * N,'value_2':[[88870.0],None, [88870.0,45.4]] * N})# 在真实数据中进行性能测试%timeit df[[a not in b if isinstance(b, list) else a != b for a, b in zip(df['value_1'], df['value_2'])]]%timeit df[[not bool(np.isin(v1, v2)) for v1, v2 in zip(df['value_1'], df['value_2'])]]
请注意，这是您提供的代码的翻译。
英文:
Use zip with list comprehension for test if lists not contains floats, if not lists are removed rows by passing False, filter in boolean indexing:
df = pd.DataFrame({'value_1':[88870.0,150700.0,392000.0],'value_2':[[88870.0],None, [88870.0,45.4]]})print (df)value_1 value_20 88870.0 [88870.0]1 150700.0 None2 392000.0 [88870.0, 45.4]mask = [a not in b if isinstance(b, list) else Falsefor a, b in zip(df['value_1'], df['value_2'])]df1 = df[mask]print (df1)value_1 value_22 392000.0 [88870.0, 45.4]
If need also test scalars:
mask = [a not in b if isinstance(b, list) else a != bfor a, b in zip(df['value_1'], df['value_2'])]df2 = df[mask]print (df2)value_1 value_21 150700.0 None2 392000.0 [88870.0, 45.4]
Performance: Pure python should be faster, best test in real data:
#20k rowsN = 10000df = pd.DataFrame({'value_1':[88870.0,150700.0,392000.0] * N,'value_2':[[88870.0],None, [88870.0,45.4]] * N})print (df)In [51]: %timeit df[[a not in b if isinstance(b, list) else a != b for a, b in zip(df['value_1'], df['value_2'])]]18.8 ms ± 1.99 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)In [52]: %timeit df[[not bool(np.isin(v1, v2)) for v1, v2 in zip(df['value_1'], df['value_2'])]]419 ms ± 3.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
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