Python Pandas 数据框获取每组的第一行
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Pandas dataframe get first row of each group
提问by Nilani Algiriyage
I have a pandas DataFramelike following.
我有一个DataFrame喜欢以下的熊猫。
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"]})
I want to group this by ["id","value"] and get the first row of each group.
我想按 ["id","value"] 对其进行分组并获取每个组的第一行。
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
Expected outcome
预期结果
id value
1 first
2 first
3 first
4 second
5 first
6 first
7 fourth
I tried following which only gives the first row of the DataFrame. Any help regarding this is appreciated.
我尝试了以下仅给出DataFrame. 对此的任何帮助表示赞赏。
In [25]: for index, row in df.iterrows():
....: df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])
采纳答案by Roman Pekar
>>> df.groupby('id').first()
value
id
1 first
2 first
3 first
4 second
5 first
6 first
7 fourth
If you need idas column:
如果您需要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
To get n first records, you can use head():
要获取第 n 个记录，您可以使用 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by wij
This will give you the second row of each group (zero indexed, nth(0) is the same as first()):
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)
Documentation: http://pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group
文档：http: //pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group
回答by Siraj S.
maybe this is what you want
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
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by vital_dml
I'd suggest to use .nth(0)rather than .first()if you need to get the first row.
如果您需要获得第一行，我建议使用.nth(0)而不是.first()。
The difference between them is how they handle NaNs, so .nth(0)will return the first row of group no matter what are the values in this row, while .first()will eventually return the first notNaNvalue in each column.
它们之间的区别在于它们如何处理 NaN，因此.nth(0)无论该行中的值是什么，都将返回 group 的第一行，而.first()最终将返回每列中的第一个notNaN值。
E.g. if your dataset is :
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
And
和
>>> df.groupby('id').first()
value
id
1 first
2 second
3 first
4 first
回答by YOBEN_S
If you only need the first row from each group we can do with drop_duplicates, Notice the function default method keep='first'.
如果您只需要我们可以使用的每个组的第一行drop_duplicates，请注意函数默认方法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
