问题:熊猫数据框获取每个组的第一行
我有DataFrame
下面的熊猫。
df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4,5,6,6,6,7,7],
'value' : ["first","second","second","first",
"second","first","third","fourth",
"fifth","second","fifth","first",
"first","second","third","fourth","fifth"]})
我想通过[“ id”,“ value”]对此分组,并获得每个分组的第一行。
id value
0 1 first
1 1 second
2 1 second
3 2 first
4 2 second
5 3 first
6 3 third
7 3 fourth
8 3 fifth
9 4 second
10 4 fifth
11 5 first
12 6 first
13 6 second
14 6 third
15 7 fourth
16 7 fifth
预期结果
id value
1 first
2 first
3 first
4 second
5 first
6 first
7 fourth
我尝试了以下操作,仅给出的第一行DataFrame
。任何有关此的帮助表示赞赏。
In [25]: for index, row in df.iterrows():
....: df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])
回答 0
>>> df.groupby('id').first()
value
id
1 first
2 first
3 first
4 second
5 first
6 first
7 fourth
如果需要id
作为列:
>>> df.groupby('id').first().reset_index()
id value
0 1 first
1 2 first
2 3 first
3 4 second
4 5 first
5 6 first
6 7 fourth
要获取n条第一条记录,可以使用head():
>>> df.groupby('id').head(2).reset_index(drop=True)
id value
0 1 first
1 1 second
2 2 first
3 2 second
4 3 first
5 3 third
6 4 second
7 4 fifth
8 5 first
9 6 first
10 6 second
11 7 fourth
12 7 fifth
回答 1
这将为您提供每组的第二行(零索引,nth(0)与first()相同):
df.groupby('id').nth(1)
文档:http : //pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group
回答 2
我建议使用.nth(0)
而不是.first()
如果您需要获得第一行。
它们之间的区别在于它们处理NaN的方式,因此.nth(0)
无论该行中的值是什么,都将返回组的第一行,而.first()
最终将返回每列中的第一个not NaN
值。
例如,如果您的数据集是:
df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4],
'value' : ["first","second","third", np.NaN,
"second","first","second","third",
"fourth","first","second"]})
>>> df.groupby('id').nth(0)
value
id
1 first
2 NaN
3 first
4 first
和
>>> df.groupby('id').first()
value
id
1 first
2 second
3 first
4 first
回答 3
也许这就是你想要的
import pandas as pd
idx = pd.MultiIndex.from_product([['state1','state2'], ['county1','county2','county3','county4']])
df = pd.DataFrame({'pop': [12,15,65,42,78,67,55,31]}, index=idx)
pop state1 county1 12 county2 15 county3 65 county4 42 state2 county1 78 county2 67 county3 55 county4 31
df.groupby(level=0, group_keys=False).apply(lambda x: x.sort_values('pop', ascending=False)).groupby(level=0).head(3)
> Out[29]:
pop
state1 county3 65
county4 42
county2 15
state2 county1 78
county2 67
county3 55
回答 4
如果只需要我们可以处理的每个组的第一行drop_duplicates
,请注意函数default方法keep='first'
。
df.drop_duplicates('id')
Out[1027]:
id value
0 1 first
3 2 first
5 3 first
9 4 second
11 5 first
12 6 first
15 7 fourth