pandas groupby 和列的每个值出现的百分比
[英]pandas groupby and percentage of occurrences of each value of a column
问题描述
I have a pandas dataframe like this and want to create a column like created_column :
我有一个像这样的熊猫数据框，想创建一个像created_column的列：
iv_1 iv_2 iv_3 iv_4 iv_5 col2rplc created_column
0 0 0 0 0 0 a 0
333 0 0 0 0 0 b 0
......
222 1 2 3 4 5 aa 1
324 1 2 3 4 5 cc 1
......
1234 1 0 0 0 1 a 1
1235 0 2 0 4 0 a 0
1236 0 0 3 0 0 a 0
1237 0 0 1 0 0 b 0
1238 0 2 0 2 0 b 0
1239 3 0 0 0 3 b 1
explanation:解释：
I want to create a column that will have 1 in rows where values in iv_5 column has occurred for less than or equal to 40% of the data, that would be for rows with values 1, 3 & 5, as shown in above example.我想创建一个列，其中iv_5列中的值发生在小于或等于 40% 的数据的行中有 1 个，这将用于值为 1、3 和 5 的行，如上例所示。 how do i do this?我该怎么做呢？
second question:第二个问题：
How do I also include less than x% and greater than y%, in creation of other column, as similar to above column creation.我如何在其他列的创建中还包含小于 x% 和大于 y% 的内容，与上面的列创建类似。
1 个解决方案
解决方案1
1 已采纳 2020-11-20 06:17:10
Use GroupBy.transform with divide length of DtaFrame and test by Series.le for less or equal:使用GroupBy.transform用的分长度DtaFrame和测试通过Series.le为小于或等于：
df['created_column'] = df.groupby('iv_5')['iv_5'].transform('size').div(len(df)).le(0.4).view('i1')
print (df)
iv_1 iv_2 iv_3 iv_4 iv_5 col2rplc created_column
0 0 0 0 0 0 a 0
333 0 0 0 0 0 b 0
222 1 2 3 4 5 aa 1
324 1 2 3 4 5 cc 1
1234 1 0 0 0 1 a 1
1235 0 2 0 4 0 a 0
1236 0 0 3 0 0 a 0
1237 0 0 1 0 0 b 0
1238 0 2 0 2 0 b 0
1239 3 0 0 0 3 b 1
Or:或者：
s = df['iv_5'].value_counts(normalize=True)
idx = s.index[s <= 0.4]
df['created_column'] = df['iv_5'].isin(idx).view('i1')
If need Series.between , both are inclusive by default, it means >= , <= , for > and < use parameter inclusive=False :如果需要Series.between ，默认情况下两者都是包含的，这意味着>= ， <= ，对于>和<使用参数inclusive=False ：
df['created_column'] = df.groupby('iv_5')['iv_5'].transform('size').div(len(df)).between(0.2, 0.5).view('i1')
print (df)
iv_1 iv_2 iv_3 iv_4 iv_5 col2rplc created_column
0 0 0 0 0 0 a 0
333 0 0 0 0 0 b 0
222 1 2 3 4 5 aa 1
324 1 2 3 4 5 cc 1
1234 1 0 0 0 1 a 0
1235 0 2 0 4 0 a 0
1236 0 0 3 0 0 a 0
1237 0 0 1 0 0 b 0
1238 0 2 0 2 0 b 0
1239 3 0 0 0 3 b 0
If need combination like > and <= between cannot be used, here is alternative:如果需要像>和<=之间的组合不能使用，这里是替代方案：
s1 = df.groupby('iv_5')['iv_5'].transform('size').div(len(df))
df['created_column'] = ((s1 > 0.2) & (s1 <= 0.6)).view('i1')
print (df)
iv_1 iv_2 iv_3 iv_4 iv_5 col2rplc created_column
0 0 0 0 0 0 a 1
333 0 0 0 0 0 b 1
222 1 2 3 4 5 aa 0
324 1 2 3 4 5 cc 0
1234 1 0 0 0 1 a 0
1235 0 2 0 4 0 a 1
1236 0 0 3 0 0 a 1
1237 0 0 1 0 0 b 1
1238 0 2 0 2 0 b 1
1239 3 0 0 0 3 b 0
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