Question

我真的被困住了。我有一个数据框，其列如下所示

Dailychange:
1
2
3
0
-1
-2
-3
1
2

我想用输出pos [3,2] nutral [1] neg [3]连续计算两个列表的正数和负数。我试过用像

这样的简单循环来解决它

    # for i in symbol:
    #     if (symbol['Dailychange']>0):
    #         counter+=1
    #         cons_list.append(counter)
    #     else:
    #         counter=0
    #         cons_list.append(counter)
    # print(cons_list)

由于我的if语句，

并输出错误。然后我尝试使用where函数

symbol['positive']=symbol.where(symbol['Dailychange']>0,'positive','Negative')

这也没有成功。我非常感谢你的帮助。

Answer 1

我们需要一个新的参数，我是使用np.where

创建的

df['New']=np.where(df['Num']>0,'positive',np.where(df['Num']==0,'Nutral','Negative'))
s=df.groupby([df['New'],(df['New']!=df['New'].shift()).cumsum()]).size().reset_index(level=1,drop=True)
s
Out[41]: 
New
Negative    3
Nutral      1
positive    3
positive    2
dtype: int64

更多信息

(df['New']!=df['New'].shift()).cumsum()
Out[804]: 
0    1
1    1
2    1
3    2
4    3
5    3
6    3
7    4
8    4
Name: New, dtype: int32

(df['New']!=df['New'].shift())
Out[805]: 
0     True 
1    False
2    False
3     True # here is the status change 
4     True # here is the status change 
5    False  # those one do not change should carry over the same number as before 
6    False
7     True # here is the status change 
8    False
Name: New, dtype: bool

我们将连续的正面或负面视为一个群体，一旦他们改变了他们的下一个群体

还有一件事True + False = 1

Answer 2

pd.cut和groupby正是你要找的 -

import numpy as np
import pandas as pd
x = pd.DataFrame([1, 2, 3, 0, -1, -2, -3, 1, 2],columns=['Dailychange'])
col = x['Dailychange']
x['Labels'] = list(pd.cut(x['Dailychange'],[-float("inf"),-0.1,0.1,float("inf")],labels=['neg','neutral','pos']))
# for i,e in enumerate(x['Labels']):
#     print(col[i],x['Labels'][i])
x['chunk_number'] = (x['Labels'] != x['Labels'].shift()).cumsum()
grouped_df = x.groupby('chunk_number')
for i in grouped_df.groups.keys():
    print(list(grouped_df.get_group(i)['Dailychange']))

同时结帐：Documentation | Related question | Another Related question

计数正数连续正数和负数

2 个答案: