pandas - 按连续范围分组 [英] Pandas - group by consecutive ranges
问题描述
我有一个具有以下结构的数据框 - 开始,结束和高度。
数据框的一些属性:
我想以一种方式对数据帧进行分组,即将高度分组为5桶长的桶,即桶分别为0,1-5,6-10,11-15和> 15 。
请参阅下面的代码示例,其中我正在寻找的是 group_by_bucket 函数的实现。
我试着寻找其他问题,但无法得到我正在寻找的确切答案。
预先感谢!
>>> d = pd.DataFrame([[1,3,5],[4,10,7],[11,17,6],[18,26,12],[27,30,15],[31, 40,6],[41,42,7]],列= ['开始','结束','高度'])
>>> d
开始结束高度
0 1 3 8
1 4 10 7
2 11 17 6
3 18 26 12
4 27 30 15
5 31 40 6
6 41 42 7
>>> d_gb = group_by_bucket(d)
>>> d_gb
start end height_grouped
0 1 17 6_10
1 18 30 11_15
2 31 42 6_10
一种方法:
df = pd.DataFrame([[1,3,10],[4,10,7],[11,17,6],[18,26,12],
[27,30,15 ],[31,40,6],[41,42,6]],columns = ['start','end','height'])
使用 cut
进行分组:
df ['groups'] = pd.cut(df.height,[ - 1,0,5,10,15,1000])
寻找断点:
df ['categories'] =(df.groups!= df.groups.shift())。cumsum()
code> df 是:
开始结束高度组类别
0 1 3 10(5,10] 0
1 4 10 7(5,10)0
2 11 17 6(5,10)0
3 18 26 12(10,15)1
4 27 30 15(10,15)1
5 31 40 6(5,10)2
6 41 42 6(5,10] 2
定义有趣的数据:
f = {'start':['first'],'end':['last'],'groups':['first']}
并使用 groupby.agg
函数: p>
df.groupby('categories')。agg(f)
groups end start $ (5,10)17 1
1(10,15)30 18
2(5,10)42 31
I have a dataframe with the following structure - Start, End and Height.
Some properties of the dataframe:
- A row in the dataframe always starts from where the previous row ended i.e. if the end for row n is 100 then the start of line n+1 is 101.
- The height of row n+1 is always different then the height in row n+1 (this is the reason the data is in different rows).
I'd like to group the dataframe in a way that heights will be grouped in buckets of 5 longs i.e. the buckets are 0, 1-5, 6-10, 11-15 and >15.
See code example below where what I'm looking for is the implemetation of group_by_bucket function.
I tried looking at other questions but couldn't get exact answer to what I was looking for.
Thanks in advance!
>>> d = pd.DataFrame([[1,3,5], [4,10,7], [11,17,6], [18,26, 12], [27,30, 15], [31,40,6], [41, 42, 7]], columns=['start','end', 'height'])
>>> d
start end height
0 1 3 8
1 4 10 7
2 11 17 6
3 18 26 12
4 27 30 15
5 31 40 6
6 41 42 7
>>> d_gb = group_by_bucket(d)
>>> d_gb
start end height_grouped
0 1 17 6_10
1 18 30 11_15
2 31 42 6_10
A way to do that :
df = pd.DataFrame([[1,3,10], [4,10,7], [11,17,6], [18,26, 12],
[27,30, 15], [31,40,6], [41, 42, 6]], columns=['start','end', 'height'])
Use cut
to make groups :
df['groups']=pd.cut(df.height,[-1,0,5,10,15,1000])
Find break points :
df['categories']=(df.groups!=df.groups.shift()).cumsum()
Then df
is :
"""
start end height groups categories
0 1 3 10 (5, 10] 0
1 4 10 7 (5, 10] 0
2 11 17 6 (5, 10] 0
3 18 26 12 (10, 15] 1
4 27 30 15 (10, 15] 1
5 31 40 6 (5, 10] 2
6 41 42 6 (5, 10] 2
"""
Define interesting data :
f = {'start':['first'],'end':['last'], 'groups':['first']}
And use the groupby.agg
function :
df.groupby('categories').agg(f)
"""
groups end start
first last first
categories
0 (5, 10] 17 1
1 (10, 15] 30 18
2 (5, 10] 42 31
"""
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