我的问题是如何用具有相同职位的列的 job_industry 值填充 nan values job_industry 列
例如：
job_title job_industry
Quality Engineer Financial Services
Quality Engineer Nan
我希望将 job_industry 的 nan 值填充为 Financial Services
如果 job_industry 中存在一个 nan 值，而 job_title 是 General manager，则用 Manufacturing 填充它
【问题讨论】：
使用groupby.ffill 和groupby.bfill 自动填充每个job_title 缺少的job_industry：
g = df.groupby('job_title')['job_industry']
df['job_industry'] = g.ffill()
df['job_industry'] = g.bfill()
# job_title job_industry
# 0 Quality Engineer Financial Services
# 1 Quality Engineer Financial Services
请注意，从技术上讲，简化的 2 行示例不需要bfill，但实际数据需要。
【讨论】：
job_title 具有 all NaN 用于 job_industry（在这种情况下，不可能知道 job_industry 的 job_title）跨度>
我将首先在job_industry 和job_title 之间创建一个映射（Python 字典），然后将job_industry 列的映射分配给job_title 的NaN 值。
代码如下：
df = pd.DataFrame(
columns=["job_title", "job_industry"],
data=[["Quality Engineer", "Financial Services"], ["Quality Engineer", None]]
)
# May be there is a faster way
title_industry_mapping = df.dropna(["job_industry"]).set_index("job_title")["job_industry"].drop_duplicates().to_dict()
isna = df["job_industry"].isna()
df.loc[isna, "job_industry"] = df.loc[isna, "job_title"].replace(title_industry_mapping)
结果：
| job_title | job_industry | |
|---|---|---|
| 0 | Quality Engineer | Financial Services |
| 1 | Quality Engineer | Financial Services |
【讨论】：
import pandas as pd
import numpy as np
df = pd.DataFrame([
['Quality Engineer','Financial Services'],
['Progammer',np.nan],
['Quality Engineer',np.nan],
['Progammer',"IT"],
['General manager',np.nan]],
columns=['job_title','job_industry'])
with pd.option_context('mode.use_inf_as_null', True):
df = df.sort_values('job_industry', ascending=False, na_position='last')
df["job_industry"].loc[(df['job_title'] == "General manager") & (df['job_industry'].isnull())] = "Manufacturing"
df['job_industry'] = df.groupby('job_title')['job_industry'].fillna(method="ffill")
df['job_industry'].isnull()，这将验证job_industry 列是否为空。
下面的代码会按照job_industry列的null值降序排序，因为如果nan值出现在前面，nan的初始值不会被替换。
with pd.option_context('mode.use_inf_as_null', True):
df = df.sort_values('job_industry', ascending=False, na_position='last')
如果你更喜欢排序而不是输出，你可以试试，df.sort_index()
O/P
+----+------------------+-------------------------------------------------------+
| | job_title | job_industry |
|----+------------------+-------------------------------------------------------|
| 0 | Quality Engineer | Financial Services |
| 1 | Progammer | IT |
| 2 | Quality Engineer | Financial Services |
| 3 | Progammer | IT |
| 4 | General manager | Manufacturing |
+----+------------------+-------------------------------------------------------+
【讨论】：
job_industry 为空，方法 2 将无法正常工作。方法3只是从我的答案中复制的。