根据要求，这是一个最小的可重现示例，它将产生 .isin() 不删除不在 .isin() 中的值而只是将它们设置为零的问题：
import os
import pandas as pd
df_example = pd.DataFrame({'Requesting as': {0: 'Employee', 1: 'Ex- Employee', 2: 'Employee', 3: 'Employee', 4: 'Ex-Employee', 5: 'Employee', 6: 'Employee', 7: 'Employee', 8: 'Ex-Employee', 9: 'Ex-Employee', 10: 'Employee', 11: 'Employee', 12: 'Ex-Employee', 13: 'Ex-Employee', 14: 'Employee', 15: 'Employee', 16: 'Employee', 17: 'Ex-Employee', 18: 'Employee', 19: 'Employee', 20: 'Ex-Employee', 21: 'Employee', 22: 'Employee', 23: 'Ex-Employee', 24: 'Employee', 25: 'Employee', 26: 'Ex-Employee', 27: 'Employee', 28: 'Employee', 29: 'Ex-Employee', 30: 'Employee', 31: 'Employee', 32: 'Ex-Employee', 33: 'Employee', 34: 'Employee', 35: 'Ex-Employee', 36: 'Employee', 37: 'Employee', 38: 'Ex-Employee', 39: 'Employee', 40: 'Employee'}, 'Years of service': {0: -0.4, 1: -0.3, 2: -0.2, 3: 1.0, 4: 1.0, 5: 1.0, 6: 2.0, 7: 2.0, 8: 2.0, 9: 2.0, 10: 3.0, 11: 3.0, 12: 3.0, 13: 4.0, 14: 4.0, 15: 4.0, 16: 5.0, 17: 5.0, 18: 5.0, 19: 5.0, 20: 6.0, 21: 6.0, 22: 6.0, 23: 11.0, 24: 11.0, 25: 11.0, 26: 16.0, 27: 17.0, 28: 18.0, 29: 21.0, 30: 22.0, 31: 23.0, 32: 26.0, 33: 27.0, 34: 28.0, 35: 31.0, 36: 32.0, 37: 33.0, 38: 35.0, 39: 36.0, 40: 37.0}, 'yos_bins': {0: 0, 1: 0, 2: 0, 3: '0-1', 4: '0-1', 5: '0-1', 6: '1-2', 7: '1-2', 8: '1-2', 9: '1-2', 10: '2-3', 11: '2-3', 12: '2-3', 13: '3-4', 14: '3-4', 15: '3-4', 16: '4-5', 17: '4-5', 18: '4-5', 19: '4-5', 20: '5-6', 21: '5-6', 22: '5-6', 23: '10-15', 24: '10-15', 25: '10-15', 26: '15-20', 27: '15-20', 28: '15-20', 29: '20-40', 30: '20-40', 31: '20-40', 32: '20-40', 33: '20-40', 34: '20-40', 35: '20-40', 36: '20-40', 37: '20-40', 38: '20-40', 39: '20-40', 40: '20-40'}})
cut_labels = ['0-1','1-2', '2-3', '3-4', '4-5', '5-6', '6-10', '10-15', '15-20', '20-40']
cut_bins = (0, 1, 2, 3, 4, 5, 6, 10, 15, 20, 40)
df_example['yos_bins'] = pd.cut(df_example['Years of service'], bins=cut_bins, labels=cut_labels)
print(df_example['yos_bins'].value_counts())
print(len(df_example['yos_bins']))
print(len(df_example))
print(df_example['yos_bins'].value_counts())
test = df_example[df_example['yos_bins'].isin(['0-1', '1-2', '2-3'])]
print('test dataframe:\n',test)
print('\n')
print('test value counts of yos_bins:\n', test['yos_bins'].value_counts())
print('\n')
dic_test = test.to_dict()
print(dic_test)
print('\n')
print(test.value_counts())ervr
我为“服务年限”的列创建了垃圾箱：
cut_labels = ['0-1','1-2', '2-3', '3-4', '4-5', '5-6', '6-10', '10-15', '15-20', '20-40']
cut_bins = (0, 1, 2, 3, 4, 5, 6, 10, 15, 20, 40)
df['yos_bins'] = pd.cut(df['Years of service'], bins=cut_bins, labels=cut_labels)
然后我将 .isin() 应用于名为“yos_bins”的数据框列，目的是过滤选择的列值。摘自 df 中的列。
我用来切片的列称为“yos_bins”（即分箱服务年限）。我只想选择 3 个范围（0-1、1-2、2-3 年），但显然该列中包含更多范围。
令我惊讶的是，当我应用 value_counts() 时，我仍然从 df 数据框中获取 yos_bins 列的所有值（但计数为 0）。
test.yos_bins.value_counts()
看起来像这样：
这不是故意的，除了 isin() 中的 3 之外的所有其他 bin 都应该被删除。产生的问题是 0 值显示在 sns.countplots 中，所以我最终得到了不想要的零计数列。
当我保存 df to_excel() 时，所有“10-15”值字段都显示“带有 2 位年份的文本日期”错误。我没有将该数据帧加载回python，所以不确定这是否会导致问题？
有谁知道我如何创建仅包含 3 个 yos_bins 值而不是显示所有 yos_bins 值但有些为零的测试数据框？