我有一个按键'按键分组的数据框。我需要比较每个组中的行,以确定是否要保留组的每一行,或者我是否只想要一组中的一行。
在保留组的所有行的条件下:如果有一行具有颜色'红色'以及' 12'的面积和“#”的形状和另一行(在同一组内),颜色为'绿色'以及' 13'和' square'的形状,然后我想保留该组中的所有行。否则,如果这种情况不存在,我想保留该组中最大的“#”数字行。值。
df = pd.DataFrame({'KEY': ['100000009', '100000009', '100000009', '100000009', '100000009','100000034','100000034', '100000034'],
'Date1': [20120506, 20120506, 20120507,20120608,20120620,20120206,20120306,20120405],
'shape': ['circle', 'square', 'circle','circle','circle','circle','circle','circle'],
'num': [3,4,5,6,7,8,9,10],
'area': [12, 13, 12,12,12,12,12,12],
'color': ['red', 'green', 'red','red','red','red','red','red']})
Date1 KEY area color num shape
0 2012-05-06 100000009 12 red 3 circle
1 2012-05-06 100000009 13 green 4 square
2 2012-05-07 100000009 12 red 5 circle
3 2012-06-08 100000009 12 red 6 circle
4 2012-06-20 100000009 12 red 7 circle
5 2012-02-06 100000034 12 red 8 circle
6 2012-03-06 100000034 12 red 9 circle
7 2012-04-05 100000034 12 red 10 circle
预期结果:
Date1 KEY area color num shape
0 2012-05-06 100000009 12 red 3 circle
1 2012-05-06 100000009 13 green 4 square
2 2012-05-07 100000009 12 red 5 circle
3 2012-06-08 100000009 12 red 6 circle
4 2012-06-20 100000009 12 red 7 circle
7 2012-04-05 100000034 12 red 10 circle
我是python的新手,而groupby给我一个曲线球。
maxnum = df.groupby('KEY')['num'].transform(max)
df = df.loc[df.num == maxnum]
cond1 = (df[df['area'] == 12]) & (df[df['color'] == 'red']) & (df[df['shape'] == 'circle'])
cond2 = (df[df['area'] == 13]) & (df[df['color'] == 'green']) & (df[df['shape'] == 'square'])
答案 0 :(得分:2)
定义名为function
的自定义函数:
def function(x):
i = x.query(
'area == 12 and color == "red" and shape == "circle"'
)
j = x.query(
'area == 13 and color == "green" and shape == "square"'
)
return x if not (i.empty or j.empty) else x[x.num == x.num.max()].head(1)
此函数在指定条件下测试每个组并根据需要返回行。特别是,它使用df.empty
查询条件并测试空虚。
将此传递给groupby
+ apply
:
df.groupby('KEY', group_keys=False).apply(function)
Date1 KEY area color num shape
0 20120506 100000009 12 red 3 circle
1 20120506 100000009 13 green 4 square
2 20120507 100000009 12 red 5 circle
3 20120608 100000009 12 red 6 circle
4 20120620 100000009 12 red 7 circle
7 20120405 100000034 12 red 10 circle