我有以下函数(这是一个模拟我在完整代码中实际操作的最小工作示例):
import pandas as pd
import numpy as np
# hardoded data for reproducibility
df = pd.DataFrame(
[
["SiteA", "Long_Key_With_KeyWord", np.nan],
["SiteA", "Long_Key_Without", np.nan],
["SiteB", "Long_Key_With_KeyWord", np.nan],
],
columns=["site", "tags", "to_fill"],
)
library = {"SiteA": {"KeyWord": "NewKeyWord"}}
# logic
df_part = df.loc[df.to_fill.isna(), :]
groupby_site = df_part.groupby("site")
for site in groupby_site.groups.keys():
site_data = groupby_site.get_group(site)
try:
library_site_data = library[site]
for idx, row in site_data.iterrows():
mask = [key in row["tags"] for key in library_site_data.keys()]
match = [key for key, mask in zip(library_site_data.keys(), mask) if mask]
if match:
value = library_site_data[match[0]]
df_part.loc[idx, "to_fill"] = value
else:
print(f"Too bad")
except KeyError:
print(f"no data for site {site} in library")
next
print(
f"Total unfound mapping tags {df.to_fill.isna().sum()}"
) # why isn't the df being filled in ?
我不明白的是为什么df
没有被填写,而我认为df_part
是指df
和df_part
被填写,它也应该填写df
我明白了:
print(df)
site tags to_fill
0 SiteA Long_Key_With_KeyWord NaN
1 SiteA Long_Key_Without NaN
2 SiteB Long_Key_With_KeyWord NaN
我想要这个:
site tags to_fill
0 SiteA Long_Key_With_KeyWord NewKeyWord
1 SiteA Long_Key_Without NaN
2 SiteB Long_Key_With_KeyWord NaN
What am I missing ?
我认为问题在于双索引(loc)可能会创建副本,而不是重新调整视图。 这个question的第一个答案进一步解释了它
相关问题 更多 >
编程相关推荐