根据列的排序更改熊猫数据框中切片的值
[英]Change values from a slice in pandas dataframe depending on the sorting of a column
问题描述
I am having some trouble in modifying values in a pandas dataframe, depending on some specific sorting.
根据某些特定的排序，我在修改熊猫数据框中的值时遇到了一些麻烦。
My DataFrame is like我的 DataFrame 就像
df_test = pd.DataFrame({'month':[1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3],
'day':[1,1,1,2,2,2,1,1,1,2,2,2,1,1,1,2,2,2],
'period':[np.random.choice(['a','b']) for i in range(18)],
'to_mark':['n']*18,
'value':np.random.randn(18)})
Where I have some months, days in that months, periods that is a categorical variable.我有几个月，那几个月的几天，这是一个分类变量的时期。 I also have a to_mak columns and a Value Column.我还有一个 to_mak 列和一个值列。 I want that, by slicing the moth, day and period column, when sorting for the value column, to change the value of 'to_mark' column from 'n' to 'y'.我希望通过切片 moth、day 和 period 列，在对 value 列进行排序时，将 'to_mark' 列的值从 'n' 更改为 'y'。
What I tried was:我尝试的是：
for m in df_test.month.unique():
for d in df_test[df_test.month==m].day.unique():
for p in df_test[(df_test.month==m) & (df_test.day==d)].period.unique():
df_test[(df_test.month==m) & (df_test.day==d) & (df_test.period == p)].sort_values(
by='value', ascending=False)['to_mark'] = 'y'
But it doesn't work properly, I am not getting to change the values of the 'to_mark' column.但它无法正常工作，我无法更改“to_mark”列的值。
One output example:一个输出示例：
Index month day period to_mark value
0 1 1 a n 0.840179
1 1 1 a n -1.349777
2 1 1 b n 0.122197
3 1 2 a n 0.276325
4 1 2 a n 0.257014
5 1 2 b n 0.351326
6 2 1 b n -0.552867
7 2 1 a n -0.614468
8 2 1 a n -0.474198
9 2 2 b n -0.439990
10 2 2 b n 0.046202
11 2 2 b n 1.601673
12 3 1 a n -1.609012
13 3 1 a n 0.382347
14 3 1 b n 0.164228
15 3 2 a n 0.176435
16 3 2 a n -0.627590
17 3 2 a n -1.834927
The desired output would be.所需的输出是。
Index month day period to_mark value
0 1 1 a y 0.840179
1 1 1 a n -1.349777
2 1 1 b y 0.122197
3 1 2 a y 0.276325
4 1 2 a n 0.257014
5 1 2 b y 0.351326
6 2 1 b n -0.552867
7 2 1 a n -0.614468
8 2 1 a y -0.474198
9 2 2 b n -0.439990
10 2 2 b n 0.046202
11 2 2 b y 1.601673
12 3 1 a n -1.609012
13 3 1 a y 0.382347
14 3 1 b y 0.164228
15 3 2 a y 0.176435
16 3 2 a n -0.627590
17 3 2 a n -1.834927
Thank you in advance.先感谢您。
1 个解决方案
解决方案1
2 已采纳 2020-02-28 13:30:09
IIUC, you just want to change the maximum for each group to 'y' ..? IIUC，您只想将每个组的最大值更改为 'y' ..？
Use idxmax , with loc to do this:使用idxmax和loc来做到这一点：
df_test = pd.DataFrame({'Index': [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17], 'month': [1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3], 'day': [1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2], 'period': ['a', 'a', 'b', 'a', 'a', 'b', 'b', 'a', 'a', 'b', 'b', 'b', 'a', 'a', 'b', 'a', 'a', 'a'], 'to_mark': ['n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n'], 'value': [0.8401790000000001, -1.349777, 0.122197, 0.276325, 0.25701399999999996, 0.351326, -0.552867, -0.614468, -0.47419799999999995, -0.43998999999999994, 0.046202, 1.601673, -1.609012, 0.382347, 0.16422799999999999, 0.17643499999999998, -0.62759, -1.834927]})
df_test.loc[df_test.groupby(['month', 'day', 'period'])['value'].idxmax(), 'to_mark'] = 'y'
[out] [出去]
Index month day period to_mark value
0 0 1 1 a y 0.840179
1 1 1 1 a n -1.349777
2 2 1 1 b y 0.122197
3 3 1 2 a y 0.276325
4 4 1 2 a n 0.257014
5 5 1 2 b y 0.351326
6 6 2 1 b y -0.552867
7 7 2 1 a n -0.614468
8 8 2 1 a y -0.474198
9 9 2 2 b n -0.439990
10 10 2 2 b n 0.046202
11 11 2 2 b y 1.601673
12 12 3 1 a n -1.609012
13 13 3 1 a y 0.382347
14 14 3 1 b y 0.164228
15 15 3 2 a y 0.176435
16 16 3 2 a n -0.627590
17 17 3 2 a n -1.834927
update更新
To update nth largest, you could use groupby with nlargest .要更新第 n 大，您可以将groupby与nlargest一起nlargest 。 Then get the indices to update by merge .:然后通过merge获取要更新的索引。：
df_test = pd.DataFrame({'month': [1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3], 'day': [1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2], 'period': ['a', 'a', 'b', 'a', 'a', 'b', 'b', 'a', 'a', 'b', 'b', 'b', 'a', 'a', 'b', 'a', 'a', 'a'], 'to_mark': ['n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n', 'n'], 'value': [0.8401790000000001, -1.349777, 0.122197, 0.276325, 0.25701399999999996, 0.351326, -0.552867, -0.614468, -0.47419799999999995, -0.43998999999999994, 0.046202, 1.601673, -1.609012, 0.382347, 0.16422799999999999, 0.17643499999999998, -0.62759, -1.834927]})
n_largest = (df_test.groupby(['month', 'day', 'period'])['value'].nlargest(2).reset_index())
idx = (df_test.reset_index()
.merge(n_largest, on=['month', 'day', 'period', 'value'],
how='inner')['index'])
df_test.loc[idx, 'to_mark'] = 'y'
[out] [出去]
month day period to_mark value
0 1 1 a y 0.840179
1 1 1 a y -1.349777
2 1 1 b y 0.122197
3 1 2 a y 0.276325
4 1 2 a y 0.257014
5 1 2 b y 0.351326
6 2 1 b y -0.552867
7 2 1 a y -0.614468
8 2 1 a y -0.474198
9 2 2 b n -0.439990
10 2 2 b y 0.046202
11 2 2 b y 1.601673
12 3 1 a y -1.609012
13 3 1 a y 0.382347
14 3 1 b y 0.164228
15 3 2 a y 0.176435
16 3 2 a y -0.627590
17 3 2 a n -1.834927
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