我正在尝试在我的 pandas 数据框中查找重复的行。实际上，df.shape 是 438796, 4531，但我使用下面的这个玩具示例来表示 MRE
| id | ft1 | ft2 | ft3 | ft4 | ft5 | label |
|:------:|:---:|:---:|:---:|:---:|:---:|:------:|
| id_100 | 1 | 1 | 43 | 1 | 1 | High |
| id_101 | 1 | 1 | 33 | 0 | 1 | Medium |
| id_102 | 1 | 1 | 12 | 1 | 1 | Low |
| id_103 | 1 | 1 | 46 | 1 | 0 | Low |
| id_104 | 1 | 1 | 10 | 1 | 1 | High |
| id_105 | 0 | 1 | 99 | 0 | 1 | Low |
| id_106 | 0 | 0 | 0 | 0 | 0 | High |
| id_107 | 1 | 1 | 6 | 0 | 1 | High |
| id_108 | 1 | 1 | 29 | 1 | 1 | Medium |
| id_109 | 1 | 0 | 27 | 0 | 0 | Medium |
| id_110 | 0 | 1 | 32 | 0 | 1 | High |
我想要完成的是观察功能的子集，如果存在重复的行，则保留第一行，然后指示哪个 id: label 对是重复的。
我查看了以下帖子:
- find duplicate rows in a pandas dataframe
(我不知道如何替换
df['index_original'] = df.groupby(['col1', 'col2']).col1.transform('idxmin' 中的和我的列列表)col1) - Find all duplicate rows in a pandas dataframe
我知道pandas有一个duplicated()称呼。所以我尝试实现它并且它有点有效:
import pandas as pd
# Read in example data
df = pd.read_clipboard()
# Declare columns I am interested in
cols = ['ft1', 'ft2', 'ft4', 'ft5']
# Create a subset of my dataframe with only the columns I care about
sub_df = df[cols]
# Create a list of duplicates
dupes = sub_df.index[sub_df.duplicated(keep='first')].tolist()
# Loop through the duplicates and print out the values I want
for idx in dupes:
# print(df[:idx])
print(df.loc[[idx],['id', 'label']])
但是，我想要做的是对于特定行，通过将这些行保存为 id: label 组合来确定哪些行是其重复项。因此，虽然我能够提取每个重复项的 id 和 label，但我无法将其映射回其重复的原始行。
理想的数据集如下所示:
| id | ft1 | ft2 | ft3 | ft4 | ft5 | label | duplicates |
|:------:|:---:|:---:|:---:|:---:|:---:|:------:|:-------------------------------------------:|
| id_100 | 1 | 1 | 43 | 1 | 1 | High | {id_102: Low, id_104: High, id_108: Medium} |
| id_101 | 1 | 1 | 33 | 0 | 1 | Medium | {id_107: High} |
| id_102 | 1 | 1 | 12 | 1 | 1 | Low | |
| id_103 | 1 | 1 | 46 | 1 | 0 | Low | |
| id_104 | 1 | 1 | 10 | 1 | 1 | High | |
| id_105 | 0 | 1 | 99 | 0 | 1 | Low | {id_110: High} |
| id_106 | 0 | 0 | 0 | 0 | 0 | High | |
| id_107 | 1 | 1 | 6 | 0 | 1 | High | |
| id_108 | 1 | 1 | 29 | 1 | 1 | Medium | |
| id_109 | 1 | 0 | 27 | 0 | 0 | Medium | |
| id_110 | 0 | 1 | 32 | 0 | 1 | High | |
如何获取重复值并将它们有效地映射回原始值(了解实际数据集的大小)？
最佳答案
在列中使用字典确实很复杂，这是一种可能的解决方案:
# Declare columns I am interested in
cols = ['ft1', 'ft2', 'ft4', 'ft5']
# Create a subset of my dataframe with only the columns I care about
sub_df = df[cols]
#mask for first dupes
m = sub_df.duplicated()
#create tuples, aggregate to list of tuples
s = (df.assign(a = df[['id','label']].apply(tuple, 1))[m]
.groupby(cols)['a']
.agg(lambda x: dict(list(x))))
#add new column
df = df.join(s.rename('duplicates'), on=cols)
#repalce missing values and not first duplciates to empty strings
df['duplicates'] = df['duplicates'].fillna('').mask(m, '')
print (df)
id ft1 ft2 ft3 ft4 ft5 label \
0 id_100 1 1 43 1 1 High
1 id_101 1 1 33 0 1 Medium
2 id_102 1 1 12 1 1 Low
3 id_103 1 1 46 1 0 Low
4 id_104 1 1 10 1 1 High
5 id_105 0 1 99 0 1 Low
6 id_106 0 0 0 0 0 High
7 id_107 1 1 6 0 1 High
8 id_108 1 1 29 1 1 Medium
9 id_109 1 0 27 0 0 Medium
10 id_110 0 1 32 0 1 High
duplicates
0 {'id_102': 'Low', 'id_104': 'High', 'id_108': ...
1 {'id_107': 'High'}
2
3
4
5 {'id_110': 'High'}
6
7
8
9
10
使用自定义函数替代，用于将没有第一个值的所有重复项分配给每组新列的第一个值，最后一个更改掩码以替换空字符串:
# Declare columns I am interested in
cols = ['ft1', 'ft2', 'ft4', 'ft5']
m = ~df.duplicated(subset=cols) & df.duplicated(subset=cols, keep=False)
def f(x):
x.loc[x.index[0], 'duplicated'] = [dict(x[['id','label']].to_numpy()[1:])]
return x
df = df.groupby(cols).apply(f)
df['duplicated'] = df['duplicated'].where(m, '')
print (df)
id ft1 ft2 ft3 ft4 ft5 label \
0 id_100 1 1 43 1 1 High
1 id_101 1 1 33 0 1 Medium
2 id_102 1 1 12 1 1 Low
3 id_103 1 1 46 1 0 Low
4 id_104 1 1 10 1 1 High
5 id_105 0 1 99 0 1 Low
6 id_106 0 0 0 0 0 High
7 id_107 1 1 6 0 1 High
8 id_108 1 1 29 1 1 Medium
9 id_109 1 0 27 0 0 Medium
10 id_110 0 1 32 0 1 High
duplicated
0 {'id_102': 'Low', 'id_104': 'High', 'id_108': ...
1 {'id_107': 'High'}
2
3
4
5 {'id_110': 'High'}
6
7
8
9
10
关于python - 使用字典将重复行映射到原始行 - Python 3.6，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/65109613/