在 for/if-else 循环中填充 np.nan 条件

Question

我已经为此工作了一段时间，但似乎无法找到我需要的答案。假设我有一个如下的数据框。

我想做的是根据df['home_work'] 列中的值填充df['gender'] 的最后三行，特别是如果home_work > 9，则m，如果不是，则f。请记住，这只是一个编造的数据集，我保证不会冒犯任何人！

enr = pd.DataFrame({'name_id':[1254, 1359, 1254, 1296, 1353, 2656], 
                   'enrollment_term':['spring 2018', 'spring 2018', 'fall 2018', 'spring 2018', 'spring 2018', 'fall 2020'],
                   'gpa_term': [2.93, np.nan, 1.65, 4.00, 3.95, 2.92],
                   'dog_owner':[0,1,1,1, 1, 0],
                   'salary':[50657, 90658, np.nan, 104352, np.nan, 102043],
                   'home_work':[34, np.nan, 12, 9, 8, 27],
                   'gender':['m','f','f',np.nan, np.nan, np.nan]})

enr

下面是我尝试的代码，但它在下面显示了错误：

for i in df['gender'].isna():
    if df['home_work'][i] > 9:
        df['gender'][i].fillna('m')
    else:
        df['gender'][i].fillna('f')

KeyError: False

任何帮助都会非常感谢，因为我已经为此工作了一段时间。我有一个 90K + 的数据集，我想调整这项工作，并想创建一个函数来简化这个过程但遇到了减速！

我遇到的问题是 np.nan 默认并填写 gender 的值，如果它不满足要求。想法？

---

#已编辑

假设我有以下df：

enr = pd.DataFrame({'name_id':[1254, 1359, 1254, 1296, 1353, 2656], 
                   'enrollment_term':['spring 2018', 'spring 2018', 'fall 2018', 'spring 2018', 'spring 2018', 'fall 2020'],
                   'gpa_term': [2.93, np.nan, 1.65, 4.00, 3.95, 2.92],
                   'dog_owner':[0,1,1,1, 1, 0],
                   'salary':[50657, 90658, np.nan, 104352, np.nan, 102043],
                   'home_work':[np.nan, np.nan, 0.7, 0.3, 0.64, 0.49],
                   'gender':[0, 1, 1,np.nan, np.nan, np.nan]})

我想根据home_work 估算enr['gender']。如果enr['home_work'] >= 0.5，则enr['gender'] == 0，否则（只要enr['home_work'] != np.nan），enr['gender'] == 1。

我不想要的是在enr[gender] 中估算值，而他们的enr['home_work'] 是np.nan我尝试了许多不同的技术，但似乎都估算为 1. 想法？

Answer 1

将numpy.where 与Series.fillna 一起使用：

enr['gender'] = np.where(enr['home_work'] > 9,  
                         enr['gender'].fillna('m'),
                         enr['gender'].fillna('f'))

或者通过2个掩码分别过滤：

m = enr['gender'].isna()
enr.loc[m, 'gender'] = np.where(enr['home_work'] > 9,  'm',  'f')[m]

---

print (enr)
   name_id enrollment_term  gpa_term  dog_owner    salary  home_work gender
0     1254     spring 2018      2.93          0   50657.0         34      m
1     1359     spring 2018       NaN          1   90658.0         42      f
2     1254       fall 2018      1.65          1       NaN         12      f
3     1296     spring 2018      4.00          1  104352.0          9      f
4     1353     spring 2018      3.95          1       NaN          8      f
5     2656       fall 2020      2.92          0  102043.0         27      m

编辑：

m = enr['gender'].isna() & enr['home_work'].notna()
enr.loc[m, 'gender'] = np.where(enr['home_work'] >= 0.5, 0, 1)[m]
print (enr)
   name_id enrollment_term  gpa_term  dog_owner    salary  home_work  gender
0     1254     spring 2018      2.93          0   50657.0        NaN     0.0
1     1359     spring 2018       NaN          1   90658.0        NaN     1.0
2     1254       fall 2018      1.65          1       NaN       0.70     1.0
3     1296     spring 2018      4.00          1  104352.0       0.30     1.0
4     1353     spring 2018      3.95          1       NaN       0.64     0.0
5     2656       fall 2020      2.92          0  102043.0       0.49     1.0

Answer 2

让我们试试map的值和where

df.gender=df.gender.where(df.gender.notna(),df.home_work.gt(9).map({True:'m',False:'f'})) 


df
   name_id enrollment_term  gpa_term  dog_owner    salary  home_work gender
0     1254     spring 2018      2.93          0   50657.0       34.0      m
1     1359     spring 2018       NaN          1   90658.0        NaN      f
2     1254       fall 2018      1.65          1       NaN       12.0      f
3     1296     spring 2018      4.00          1  104352.0        9.0      f
4     1353     spring 2018      3.95          1       NaN        8.0      f
5     2656       fall 2020      2.92          0  102043.0       27.0      m