Question

df = pd.DataFrame({'Campaign ID':[48464,48464,48464,48464,26380,26380,22676,39529,39529,46029,46029,46029,17030,46724,46724,39379,39379,39379],
'Campaign stage':["Lost","Developing","Discussing","Starting","Discussing", "Starting","Developing",    "Discussing","Starting","Developing",   "Discussing","Starting","Developing",   "Developing","Discussing","Lost",   "Developing","Discussing"],
'Stage Number':[-1, 3,  2,  1,  2,  1,  3,  2,  1,  3,  2,  1,  3,  3,  2,  -1, 3,  2],
'Campaign Date':["2/8/2019","1/9/2019","1/3/2019","3/3/2018","2/14/2019","12/5/2018","7/25/2018","6/8/2018","3/4/2018","12/8/2018","9/9/2018","5/31/2018","6/7/2018","3/27/2018","1/6/2018","2/15/2019","12/15/2018","9/4/2018"]})

pvt = pd.pivot_table(df,values=['Campaign stage'],index=['Campaign ID','Campaign stage','Stage Number','Campaign Date'],aggfunc='count')
pvt.sort_values(['Campaign ID','Campaign Date'],ascending=[True,False])

大家好，我具有上面的数据框，我想计算每个广告系列在广告系列“开始”和“讨论”之间的天数，然后计算平均值。

由于数据质量，活动阶段不一致。因此，对于广告系列没有“开始”和“讨论”两个阶段的情况，我想将其设置为0。

我创建了数据的数据透视表视图，并对广告系列日期的降序进行了排序...但是我不知道下一步该怎么做。

预先感谢您的帮助。

Answer 1

这是我的建议，从您的df开始：

df['Campaign Date'] = pd.to_datetime(df['Campaign Date'])
df = df[df['Campaign stage'].isin(['Starting', 'Discussing'])]

pvt = pd.pivot_table(df,values=['Campaign stage'],index=['Campaign ID','Campaign stage','Stage Number','Campaign Date'],aggfunc='count')
pvt = pvt.sort_values(['Campaign ID','Campaign Date'],ascending=True).reset_index(level=3)

pvt.groupby(level=0).diff().sum(level=0)

输出为：

Campaign ID - Campaign Date
26380   71 days
39379   0 days
39529   96 days
46029   101 days
46724   0 days
48464   306 days

如果您要查找的是总均值：

pvt.groupby(level=0).diff().sum(level=0).mean()

哪个是

95 days 16:00:00

Answer 2

我的解决方案消除了对表格视图的需求。

df['Campaign Date'] = pd.to_datetime(df['Campaign Date'])
days = []
for ID in df['Campaign ID'].unique():
    try:
        a = df[(df['Campaign ID'] == ID) & (df['Campaign stage'] == "Starting")].iloc[0]["Campaign Date"]
        b = df[(df['Campaign ID'] == ID) & (df['Campaign stage'] == "Discussing")].iloc[0]["Campaign Date"]
        days.append((b - a).days)
    except:
        days.append(0)

average = sum(days) / len(days)

Answer 3

df['Campaign Date'] =  pd.to_datetime(df['Campaign Date'],format='%m/%d/%Y')
compare= {}
for ids,gp in df.groupby('Campaign ID'):
    try:
        compare[ids]= gp.loc[gp['Campaign stage']=='Discussing']['Campaign Date'].iloc[0] - gp.loc[gp['Campaign stage']=='Starting']['Campaign Date'].iloc[0]
    except:
        compare[ids] =0

df['new_col'] = df['Campaign ID'].apply(lambda x:compare[x])

Answer 4

我不确定您希望最终结果采用哪种形式，下面的示例假定您希望在新的数据框中使用它。

不过，我们可以走多条路线，但是我选择在下面创建一个函数，然后为了清楚起见而分解每个步骤。

第1步：将您的日期数据类型更新为datetime进行比较
第2步：调用函数第2.1步：定义一个空列表来存储我们的cid值和starting和discussing之间的天数
- 步骤2.2：获取所有唯一的cid并遍历每个
- 第2.3步：检查cid是否同时具有我们感兴趣的两个阶段
  - 步骤2.3.1：如果未为此cid分配0
  - 步骤2.3.2：否则计算discussing和starting之间的差异
- 步骤2.4：创建我们的新数据框并返回
- 步骤2.5：使用df.loc创建并计算我们的平均日索引

代码

# First we need to modify your data type to compare the dates
df['Campaign Date'] = pd.to_datetime(df['Campaign Date'])

df_days = getDays(df)

# We create a new index called 'Average' that will assign values to the days column
df_days.loc['Average','days'] = df_days.days.mean()

def getDays(df):
    days = list()
    cids = list()

    for cid in df['Campaign ID'].unique():
        stages = df[(df['Campaign ID'] == cid) & 
                    (df['Campaign stage'].str.lower().isin(['discussing', 'starting']))]['Campaign stage'].unique()

        if len(stages) < 2:
            d = 0
            c = cid

            days.append(d)
            cids.append(c)

        else:
            starting = df[(df['Campaign ID'] == cid) & (df['Campaign stage'].str.lower() == 'starting')]['Campaign Date'].dt.date.values
            discussing = df[(df['Campaign ID'] == cid) & (df['Campaign stage'].str.lower() == 'discussing')]['Campaign Date'].dt.date.values

            d = (discussing - starting)[0].days

            c = cid

            days.append(d)
            cids.append(c)


    dff = pd.DataFrame({
        'cids': cids,
        'days': days
    })          

    return dff

退出

        cids    days
0       48464.0 306.00
1       26380.0 71.00
2       22676.0 0.00
3       39529.0 96.00
4       46029.0 101.00
5       17030.0 0.00
6       46724.0 0.00
7       39379.0 0.00
Average NaN   71.75

如何计算此df Python熊猫在各个阶段之间的天数？

4 个答案: