我有一个列表,如
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
和一个数据框,如
A 100
B 200
C 300
D 400
我想从上面的列表中得到一个小组总和成为:
Group 1 300
Group 2 700
我怎么能用python pandas做到这一点? 不用说我是熊猫的新手。感谢。
答案 0 :(得分:4)
您需要按$('#calendar').fullCalendar({
header: {
left: 'prev,next today',
center: 'title',
right: 'month,agendaWeek,agendaDay'
},
events: _this.state.events,
defaultView:'month',
displayEventTime: false,
editable: false,
droppable: false,
durationEditable: false
});
然后groupby
创建viewRender: function(view, element) {
//note: this is a hack, i don't know why the view title keep showing "undefined" text in it.
//probably bugs in jquery fullcalendar
$('.fc-center')[0].children[0].innerText = view.title.replace(new RegExp("undefined", 'g'), ""); ;
},
并汇总[{"start":"2017-03-24T00:00:00.000Z","end":"2017-03-26T00:00:00.000Z","title":"Open house","description":"Bali 1 open house"}]
:
dict
可能是一点修改解决方案 - 如果只有lists
列由sum
聚合。最后reset_index
用于将索引转换为列。
df = pd.DataFrame({'a': ['A', 'B', 'C', 'D'], 'b': [100, 200, 300, 400]})
print (df)
a b
0 A 100
1 B 200
2 C 300
3 D 400
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
#http://stackoverflow.com/q/43227103/2901002
d = {k:row[0] for row in groups for k in row[1:]}
print (d)
{'B': 'Group1', 'C': 'Group2', 'D': 'Group2', 'A': 'Group1'}
print (df.set_index('a').groupby(d).sum())
b
Group1 300
Group2 700
答案 1 :(得分:1)
另一种选择......但似乎@ jezrael的方式更好!
import pandas as pd
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
df0 = pd.melt(pd.DataFrame(groups).set_index(0).T)
df1 = pd.read_clipboard(header=None) # Your example data
df = df1.merge(df0, left_on=0, right_on='value')[['0_y', 1]]
df.columns = ['Group', 'Value']
print df.groupby('Group').sum()
Value
Group
Group1 300
Group2 700
答案 2 :(得分:1)
使用python 3解包和理解来创建字典。在第一列的地图中使用该字典。使用该映射分组。
考虑列表groups
和数据框df
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
df = pd.DataFrame(dict(a=list('ABCD'), b=range(100, 401, 100)))
然后:
df.groupby(df.a.map({k: g for g, *c in groups for k in c})).sum()
b
a
Group1 300
Group2 700