Python 使用np.where定义多个条件句
我试图将几个相对简单的条件组合成一个np.where子句,但是我很难理解逻辑的语法 我当前的数据帧看起来像下面的df,有四列。我想添加两列,命名如下,条件如下: 所需输出低于-dfdf_so_v2Python 使用np.where定义多个条件句,python,pandas,numpy,dataframe,Python,Pandas,Numpy,Dataframe,我试图将几个相对简单的条件组合成一个np.where子句,但是我很难理解逻辑的语法 我当前的数据帧看起来像下面的df,有四列。我想添加两列,命名如下,条件如下: 所需输出低于-dfdf_so_v2 活动后天数 *查找具有相同ID的上一行的最近一行,然后减去日期列 *如果没有最新值,则返回NA Chg。平均值 条件1:如果计数为0,则不适用 条件2:如果计数=0,查找具有相同ID和计数的最近的上一行=0,然后在“平均值”列中查找差异 然而,我正在构建一个简单的np.where查询,如下所示,我不知
df['CASH'] = np.where(df['CASH'] != 0, df['CASH'] + commission , df['CASH'])
非常感谢你在这方面的帮助
df_dict={'DateOf': ['2017-08-07','2017-08-07','2017-08-07','2017-08-04','2017-08-04','2017-08-04'
, '2017-08-03','2017-08-03','2017-08-03','2017-08-02','2017-08-02','2017-08-02','2017-08-01','2017-08-01','2017-08-01'],
'ID': ['553','559','914','553','559','914','553','559','914','553','559','914','553','559','914'], 'Count': [0, 4, 5, 0, 11, 10, 3, 9, 0,1,0,2,4,4,0],
'Avg. Value': [0,3.5,2.2,0,4.2,3.3,5.3,5,0,3,0,2,4.4,6.4,0]}
df_so=pd.DataFrame(df_dict)
df_dict_v2={'DateOf': ['2017-08-07','2017-08-07','2017-08-07','2017-08-04','2017-08-04','2017-08-04'
, '2017-08-03','2017-08-03','2017-08-03','2017-08-02','2017-08-02','2017-08-02','2017-08-01','2017-08-01','2017-08-01'],
'ID': ['553','559','914','553','559','914','553','559','914','553','559','914','553','559','914'], 'Count': [0, 4, 5, 0, 11, 10, 3, 9, 0,1,0,2,4,4,0],
'Avg. Value': [0,3.5,2.2,0,4.2,3.3,5.3,5,0,3,0,2,4.4,6.4,0],
'Days_since_activity': [4,3,1,1,1,2,1,2,1,1,1,1,'NA','NA','NA'],
'Chg. Avg Value': ['NA',-0.7,-1.1,'NA',-0.8,1.3,2.3,-1.4,'NA',-1.4,'NA','NA','NA','NA','NA']
}
df_so_v2=pd.DataFrame(df_dict_v2)
这是问题这一部分的答案。我需要更多关于2的条件的澄清 1) 自活动开始的天数*查找具有相同ID的上一行的最近日期,然后减去日期列*如果没有最新值,则返回NA 首先需要将字符串转换为datetime,然后按升序对日期进行排序。最后使用
.transform
查找差异
df_dict={'DateOf': ['2017-08-07','2017-08-07','2017-08-07','2017-08-04','2017-08-04','2017-08-04'
, '2017-08-03','2017-08-03','2017-08-03','2017-08-02','2017-08-02','2017-08-02','2017-08-01','2017-08-01','2017-08-01'],
'ID': ['553','559','914','553','559','914','553','559','914','553','559','914','553','559','914'], 'Count': [0, 4, 5, 0, 11, 10, 3, 9, 0,1,0,2,4,4,0],
'Avg. Value': [0,3.5,2.2,0,4.2,3.3,5.3,5,0,3,0,2,4.4,6.4,0]}
df_so = pd.DataFrame(df_dict)
df_so['DateOf'] = pd.to_datetime(df_so['DateOf'])
df_so.sort_values('DateOf', inplace=True)
df_so['Days_since_activity'] = df_so.groupby(['ID'])['DateOf'].transform(pd.Series.diff)
df_so.sort_index()
df_dict={'DateOf': ['2017-08-07','2017-08-07','2017-08-07','2017-08-04','2017-08-04','2017-08-04'
, '2017-08-03','2017-08-03','2017-08-03','2017-08-02','2017-08-02','2017-08-02','2017-08-01','2017-08-01','2017-08-01'],
'ID': ['553','559','914','553','559','914','553','559','914','553','559','914','553','559','914'], 'Count': [0, 4, 5, 0, 11, 10, 3, 9, 0,1,0,2,4,4,0],
'Avg. Value': [0,3.5,2.2,0,4.2,3.3,5.3,5,0,3,0,2,4.4,6.4,0]}
df = pd.DataFrame(df_dict)
df['DateOf'] = pd.to_datetime(df['DateOf'], format='%Y-%m-%d')
df.sort_values(['ID','DateOf'], inplace=True)
df['Days_since_activity'] = df.groupby(['ID'])['DateOf'].diff()
mask = df.ID != df.ID.shift(1)
mask2 = df.groupby('ID').Count.shift(1) == 0
df['Days_since_activity'][mask] = np.nan
df['Days_since_activity'][mask2] = df.groupby(['ID'])['DateOf'].diff(2)
df['Chg. Avg Value'] = df.groupby(['ID'])['Avg. Value'].diff()
df['Chg. Avg Value'][mask2] = df.groupby(['ID'])['Avg. Value'].diff(2)
conditions = [((df['Count'] == 0)),]
choices = [np.nan,]
df['Chg. Avg Value'] = np.select(conditions, choices, default = df['Chg. Avg Value'])
# df = df.sort_index()
df
根据您的评论编辑:
查找计数不为零的最近前一天,并计算差值
df_dict={'DateOf': ['2017-08-07','2017-08-07','2017-08-07','2017-08-04','2017-08-04','2017-08-04'
, '2017-08-03','2017-08-03','2017-08-03','2017-08-02','2017-08-02','2017-08-02','2017-08-01','2017-08-01','2017-08-01'],
'ID': ['553','559','914','553','559','914','553','559','914','553','559','914','553','559','914'], 'Count': [0, 4, 5, 0, 11, 10, 3, 9, 0,1,0,2,4,4,0],
'Avg. Value': [0,3.5,2.2,0,4.2,3.3,5.3,5,0,3,0,2,4.4,6.4,0]}
df_so = pd.DataFrame(df_dict)
df_so['DateOf'] = pd.to_datetime(df_so['DateOf'])
df_so.sort_values('DateOf', inplace=True)
df_so['Days_since_activity'] = df_so.groupby(['ID'])['DateOf'].transform(pd.Series.diff)
df_so.sort_index()
df_dict={'DateOf': ['2017-08-07','2017-08-07','2017-08-07','2017-08-04','2017-08-04','2017-08-04'
, '2017-08-03','2017-08-03','2017-08-03','2017-08-02','2017-08-02','2017-08-02','2017-08-01','2017-08-01','2017-08-01'],
'ID': ['553','559','914','553','559','914','553','559','914','553','559','914','553','559','914'], 'Count': [0, 4, 5, 0, 11, 10, 3, 9, 0,1,0,2,4,4,0],
'Avg. Value': [0,3.5,2.2,0,4.2,3.3,5.3,5,0,3,0,2,4.4,6.4,0]}
df = pd.DataFrame(df_dict)
df['DateOf'] = pd.to_datetime(df['DateOf'], format='%Y-%m-%d')
df.sort_values(['ID','DateOf'], inplace=True)
df['Days_since_activity'] = df.groupby(['ID'])['DateOf'].diff()
mask = df.ID != df.ID.shift(1)
mask2 = df.groupby('ID').Count.shift(1) == 0
df['Days_since_activity'][mask] = np.nan
df['Days_since_activity'][mask2] = df.groupby(['ID'])['DateOf'].diff(2)
df['Chg. Avg Value'] = df.groupby(['ID'])['Avg. Value'].diff()
df['Chg. Avg Value'][mask2] = df.groupby(['ID'])['Avg. Value'].diff(2)
conditions = [((df['Count'] == 0)),]
choices = [np.nan,]
df['Chg. Avg Value'] = np.select(conditions, choices, default = df['Chg. Avg Value'])
# df = df.sort_index()
df
新的未排序输出便于比较:
DateOf ID Count Avg. Value Days_since_activity Chg. Avg Value
12 2017-08-01 553 4 4.4 NaT NaN
9 2017-08-02 553 1 3.0 1 days -1.4
6 2017-08-03 553 3 5.3 1 days 2.3
3 2017-08-04 553 0 0.0 1 days NaN
0 2017-08-07 553 0 0.0 4 days NaN
13 2017-08-01 559 4 6.4 NaT NaN
10 2017-08-02 559 0 0.0 1 days NaN
7 2017-08-03 559 9 5.0 2 days -1.4
4 2017-08-04 559 11 4.2 1 days -0.8
1 2017-08-07 559 4 3.5 3 days -0.7
14 2017-08-01 914 0 0.0 NaT NaN
11 2017-08-02 914 2 2.0 NaT NaN
8 2017-08-03 914 0 0.0 1 days NaN
5 2017-08-04 914 10 3.3 2 days 1.3
2 2017-08-07 914 5 2.2 3 days -1.1
索引11应该是NAT,因为当前最前面的行有0个计数,没有别的东西可以与
比较。你能张贴一些想要的输出吗?考虑使用<代码> NP。选择< /代码> @黑暗NP-选择是这个ChRISZ的方法,所需的输出是DfySoV2OK。首先使用np.select添加条件,然后进行实际的减法运算?谢谢,克里斯。我将对第2部分作一些澄清。但是,对于您的回答,它不考虑有关计数的条件。例如,对于第#0行,由于前一天该ID的计数为零,因此“自#u”活动的天数应返回4天。这有意义吗?