Question

我有一张下表。第一列是年份，第二列是路面处理类型，第三列是路面得分。我需要通过从当前分数的年份中减去最后一次处理的年份来创建第三列，称为“ year diff”。例如，2014年需要减去2013，因为处理9是在2013年完成的，结果1必须记录在相应单元格的col ['year diff']中。由于处理10是在2020年完成的，因此2022年需要减去2020。

非常感谢大家的帮助。

真诚的

威尔逊

Answer 1

使用：

#check not missing values
m = df['treatment'].notnull()
#create groups starting not missing values
s = m.cumsum()
#add missing values for first group and for not missing values
mask = (s == 0) | m

#subtract score with first score per group 
out =  df['score'] - df['score'].groupby(s).transform('first')
#add missing values
df['year diff'] = np.where(mask, np.nan, out)
print (df)
    year  treatment  score  year diff
0   2010        NaN      1        NaN
1   2011        NaN      2        NaN
2   2012        NaN      3        NaN
3   2013        9.0      4        NaN
4   2014        NaN      5        1.0
5   2015        NaN      6        2.0
6   2016        NaN      7        3.0
7   2017        NaN      8        4.0
8   2018        NaN      9        5.0
9   2019        NaN     10        6.0
10  2020       10.0     11        NaN
11  2021        NaN     12        1.0
12  2022        NaN     13        2.0
13  2023        NaN     14        3.0
14  2024        NaN     15        4.0
15  2025       12.0     16        NaN
16  2026        NaN     17        1.0
17  2027        NaN     18        2.0

Answer 2

IIUC，您可以使用：

df['identifier']=(df['year'].diff().eq(1)&df['treatment'].notnull()).cumsum()
df['year diff ']=df.groupby('identifier')['identifier'].apply\
(lambda x: pd.Series(np.where(x!=0,pd.Series(pd.factorize(x)[0]+1).cumsum().shift(),np.nan))).values
print(df)

或者如果您需要根据治疗值考虑分数差异：

df['identifier']=(df['year'].diff().eq(1) &df['treatment'].notnull()).cumsum()
df['year diff']=df.groupby('identifier')['score']\
.apply(lambda x : pd.Series(np.where(x!=0,x.diff().expanding().sum(),np.nan))).reset_index(drop=True)
df.loc[df['identifier']==0,'year diff']=np.nan
print(df)

    year  treatment  score  identifier  year diff 
0   2010        NaN      1           0         NaN
1   2011        NaN      2           0         NaN
2   2012        NaN      3           0         NaN
3   2013        9.0      4           1         NaN
4   2014        NaN      5           1         1.0
5   2015        NaN      6           1         2.0
6   2016        NaN      7           1         3.0
7   2017        NaN      8           1         4.0
8   2018        NaN      9           1         5.0
9   2019        NaN     10           1         6.0
10  2020       10.0     11           2         NaN
11  2021        NaN     12           2         1.0
12  2022        NaN     13           2         2.0
13  2023        NaN     14           2         3.0
14  2024        NaN     15           2         4.0
15  2025       12.0     16           3         NaN
16  2026        NaN     17           3         1.0
17  2027        NaN     18           3         2.0

Answer 3

如果您想使用for循环来完成此操作：

df = pd.DataFrame(mydata) 
mylist = df.index[df['treatment'] != ''].tolist()

现在我们减去year值

re_list= []
for index,row in df.iterrows():
    if index > min(mylist):
        m = [i for i in mylist if i <= index]
        re_list.append(df.iloc[index]['year'] - df.iloc[max(m)]['year'])
else:
    re_list.append(0)

df['Result'] = re_list

在熊猫的两个专栏之间的年差异

3 个答案: