I have a df like this,
我有这样的df,
df,
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
df_mask = pd.DataFrame({'AAA' : [True] * 4, 'BBB' : [False] * 4,'CCC' : [True,False] * 2})
and df.where(df_mask)
is
df_mask = pd.DataFrame({'AAA':[True] * 4,'BBB':[False] * 4,'CCC':[True,False] * 2})和df.where(df_mask)是
AAA BBB CCC
0 4 NaN 100.0
1 5 NaN NaN
2 6 NaN -30.0
3 7 NaN NaN
I am trying to extract the non null values like this.
我试图像这样提取非空值。
I tried, df[df.where(df_mask).notnull()].to_dict()
but it gives all the values
我试过,df [df.where(df_mask).notnull()]。to_dict()但它给出了所有的值
My expected output is,
我的预期产量是,
{'AAA': {0: 4, 1: 5, 2: 6, 3: 7},
'CCC': {0: 100.0, 2: -30.0}}
2 个解决方案
#1
2
Let's use agg
here:
我们在这里使用agg:
v = df.where(df_mask).agg(lambda x: x.dropna().to_dict())
On older versions, apply
does the same thing (albeit a bit slower).
在旧版本中,apply会做同样的事情(虽然有点慢)。
v = df.where(df_mask).apply(lambda x: x.dropna().to_dict())
And now, filter out rows with empty dictionaries for the final step:
现在,为最后一步过滤掉带有空字典的行:
res = v[v.str.len() > 0].to_dict()
print(res)
{'AAA': {0: 4.0, 1: 5.0, 2: 6.0, 3: 7.0}, 'CCC': {0: 100.0, 2: -30.0}}
Another apply-free option is a dict-comprehension:
另一个免费申请是字典理解:
v = df.where(df_mask)
res = {k : v[k].dropna().to_dict() for k in df}
print(res)
{'AAA': {0: 4, 1: 5, 2: 6, 3: 7}, 'BBB': {}, 'CCC': {0: 100.0, 2: -30.0}}
Note that this (slightly) simpler solution retains keys with empty values.
请注意,这个(稍微)更简单的解决方案会保留具有空值的键。
#2
1
You can iterate df
's columns and apply dropna
Series
wise
您可以迭代df的列并应用dropna Serieswise
{col: df[col].dropna().values for col in df}
Which yields
{'AAA': array([4, 5, 6, 7]),
'BBB': array([], dtype=float64),
'CCC': array([ 100., -30.])}
You can filter out empty arrays such as 'BBB'
with
您可以过滤掉空数组,例如'BBB'
{key: val for key, val in ddict.items() if val}
#1
2
Let's use agg
here:
我们在这里使用agg:
v = df.where(df_mask).agg(lambda x: x.dropna().to_dict())
On older versions, apply
does the same thing (albeit a bit slower).
在旧版本中,apply会做同样的事情(虽然有点慢)。
v = df.where(df_mask).apply(lambda x: x.dropna().to_dict())
And now, filter out rows with empty dictionaries for the final step:
现在,为最后一步过滤掉带有空字典的行:
res = v[v.str.len() > 0].to_dict()
print(res)
{'AAA': {0: 4.0, 1: 5.0, 2: 6.0, 3: 7.0}, 'CCC': {0: 100.0, 2: -30.0}}
Another apply-free option is a dict-comprehension:
另一个免费申请是字典理解:
v = df.where(df_mask)
res = {k : v[k].dropna().to_dict() for k in df}
print(res)
{'AAA': {0: 4, 1: 5, 2: 6, 3: 7}, 'BBB': {}, 'CCC': {0: 100.0, 2: -30.0}}
Note that this (slightly) simpler solution retains keys with empty values.
请注意,这个(稍微)更简单的解决方案会保留具有空值的键。
#2
1
You can iterate df
's columns and apply dropna
Series
wise
您可以迭代df的列并应用dropna Serieswise
{col: df[col].dropna().values for col in df}
Which yields
{'AAA': array([4, 5, 6, 7]),
'BBB': array([], dtype=float64),
'CCC': array([ 100., -30.])}
You can filter out empty arrays such as 'BBB'
with
您可以过滤掉空数组,例如'BBB'
{key: val for key, val in ddict.items() if val}