下午全部,
我一直试图解决这个问题,任何帮助都会受到赞赏。
这是我的数据框:
Channel state rfq_qty
A Done 10
B Tied Done 10
C Done 10
C Done 10
C Done 10
C Tied Done 10
B Done 10
B Done 10
我想:
- 按渠道分组,然后是州
- 汇总每个频道的rfq_qty
- 计算状态中每个'完成'字符串的出现次数('完成'与'绑定完成'相同,即其中包含'完成'的任何内容)
- 将频道rfq_qty显示为rfq_qty(80)总数的百分比
醇>
Channel state rfq_qty Percentage
A 1 10 0.125
B 3 30 0.375
C 4 40 0.5
我尝试了以下内容:
df_Done = df[
(
df['state']=='Done'
)
|
(
df['state'] == 'Tied Done'
)
][['Channel','state','rfq_qty']]
df_Done['Percentage_Qty']= df_Done['rfq_qty']/df_Done['rfq_qty'].sum()
df_Done['Done_Trades']= df_Done['state'].count()
display(
df_Done[
(df_Done['Channel'] != 0)
].groupby(['Channel'])['Channel','Count of Done','rfq_qty','Percentage_Qty'].sum().sort_values(['rfq_qty'], ascending=False)
)
工作但看起来很复杂。有什么改进吗?
答案 0 :(得分:1)
我认为你可以使用:
isin
和loc
groupby
并按agg
汇总新列名称和函数的元组div
和Percentage
sum
sort_values
rfq_qty
df_Done = df.loc[df['state'].isin(['Done', 'Tied Done']), ['Channel','state','rfq_qty']]
#if want filter all values contains Done
#df_Done = df[df['state'].str.contains('Done')]
#if necessary filter out Channel == 0
#mask = (df['Channel'] != 0) & df['state'].isin(['Done', 'Tied Done'])
#df_Done = df.loc[mask, ['Channel','state','rfq_qty']]
d = {('rfq_qty', 'sum'), ('Done_Trades','size')}
df = df_Done.groupby('Channel')['rfq_qty'].agg(d).reset_index()
df['Percentage'] = df['rfq_qty'].div(df['rfq_qty'].sum())
df = df.sort_values('rfq_qty')
print (df)
Channel Done_Trades rfq_qty Percentage
0 A 1 10 0.125
1 B 3 30 0.375
2 C 4 40 0.500
答案 1 :(得分:0)
一种方法是使用单个df.groupby.agg
并重命名列:
import pandas as pd
df = pd.DataFrame({'Channel': ['A', 'B', 'C', 'C', 'C', 'C', 'B', 'B'],
'state': ['Done', 'Tied Done', 'Done', 'Done', 'Done', 'Tied Done', 'Done', 'Done'],
'rfq_qty': [10, 10, 10, 10, 10, 10, 10, 10]})
agg_funcs = {'state': lambda x: x[x.str.contains('Done')].count(),
'rfq_qty': ['sum', lambda x: x.sum() / df['rfq_qty'].sum()]}
res = df.groupby('Channel').agg(agg_funcs).reset_index()
res.columns = ['Channel', 'state', 'rfq_qty', 'Percentage']
# Channel state rfq_qty Percentage
# 0 A 1 10 0.125
# 1 B 3 30 0.375
# 2 C 4 40 0.500
这不是最有效的方式,因为它依赖于非向量化聚合,但如果它符合您的用例,则可能是一个不错的选择。