I have a dictionary of Pandas Series objects that I want to turn into a Dataframe. The key for each series should be the column heading. The individual series overlap but, each label is unique.
I thought I should be able to just do
df = pd.DataFrame(data)
But I keep getting the error InvalidIndexError: Reindexing only valid with uniquely valued Index objects.
I get the same error if I try to turn each series into a frame, and use pd.concat(data, axis=1).
Which doesn't make sense if you take the column label into account. What am I doing wrong, and how do I fix it?
I believe you need reset_index with parameter drop=True of each Series in dict comprehension, because duplicates in index:
s = pd.Series([1,4,5,2,0], index=[1,2,2,3,5])
s1 = pd.Series([5,7,8,1],index=[1,2,3,4])
data = {'a':s, 'b': s1}
print (s.reset_index(drop=True))
0 1
1 4
2 5
3 2
4 0
dtype: int64
df = pd.concat({k:v.reset_index(drop=True) for k,v in data.items()}, axis=1)
print (df)
a b
0 1 5.0
1 4 7.0
2 5 8.0
3 2 1.0
4 0 NaN
If you need drop rows where duplicated index use boolean indexing with duplicated:
print (s[~s.index.duplicated()])
1 1
2 4
3 2
5 0
dtype: int64
df = pd.concat({k:v[~v.index.duplicated()] for k,v in data.items()}, axis=1)
print (df)
a b
1 1.0 5.0
2 4.0 7.0
3 2.0 8.0
4 NaN 1.0
5 0.0 NaN
Another solution:
print (s.groupby(level=0).mean())
1 1.0
2 4.5
3 2.0
5 0.0
dtype: float64
df = pd.concat({k:v.groupby(level=0).mean() for k,v in data.items()}, axis=1)
print (df)
a b
1 1.0 5.0
2 4.5 7.0
3 2.0 8.0
4 NaN 1.0
5 0.0 NaN
Related
I have a main dataframe and a sub dataframe. I want to merge each column in sub dataframe into main dataframe with main dataframe column as a reference. I have successfully arrived at my desired answer, except that I see duplicated columns of the main dataframe. Below are the my expected and present answers.
Present solution:
df = pd.DataFrame({'Ref':[1,2,3,4]})
df1 = pd.DataFrame({'A':[2,3],'Z':[1,2]})
df = [df.merge(df1[col_name],left_on='Ref',right_on=col_name,how='left') for col_name in df1.columns]
df = pd.concat(df,axis=1)
df =
Ref A Ref Z
0 1 NaN 1 1.0
1 2 2.0 2 2.0
2 3 3.0 3 NaN
3 4 NaN 4 NaN
Expected Answer:
df =
Ref A Z
0 1 NaN 1.0
1 2 2.0 2.0
2 3 3.0 NaN
3 4 NaN NaN
Update
Use duplicated:
>>> df.loc[:, ~df.columns.duplicated()]
Ref A Z
0 1 NaN 1.0
1 2 2.0 2.0
2 3 3.0 NaN
3 4 NaN NaN
Old answer
You can use:
# Your code
...
df = pd.concat(df, axis=1)
# Use pop and insert to cleanup your dataframe
df.insert(0, 'Ref', df.pop('Ref').iloc[:, 0])
Output:
>>> df
Ref A Z
0 1 NaN 1.0
1 2 2.0 2.0
2 3 3.0 NaN
3 4 NaN NaN
What about setting 'Ref' col as index while getting dataframe list. (And resetting index such that you get back Ref as a column)
df = pd.DataFrame({'Ref':[1,2,3,4]})
df1 = pd.DataFrame({'A':[2,3],'Z':[1,2]})
df = [df.merge(df1[col_name],left_on='Ref',right_on=col_name,how='left').set_index('Ref') for col_name in df1.columns]
df = pd.concat(df,axis=1)
df = df.reset_index()
Ref A Z
1 NaN 1.0
2 2.0 2.0
3 3.0 NaN
4 NaN NaN
This is a reduction process. Instead of the list comprehension use for - loop, or even reduce:
from functools import reduce
reduce(lambda x, y : x.merge(df1[y],left_on='Ref',right_on=y,how='left'), df1.columns, df)
Ref A Z
0 1 NaN 1.0
1 2 2.0 2.0
2 3 3.0 NaN
3 4 NaN NaN
The above is similar to:
for y in df1.columns:
df = df.merge(df1[y],left_on='Ref',right_on=y,how='left')
df
Ref A Z
0 1 NaN 1.0
1 2 2.0 2.0
2 3 3.0 NaN
3 4 NaN NaN
I have the following two pandas dataframes:
df1
A B C
0 1 2 1
1 7 3 6
2 3 10 11
df2
A B C
0 2 0 2
1 8 4 7
Where A,B and C are column headings of both dataframes.
I am trying to compare columns of df1 to columns of df2 such that the first row in df2 is the lower bound and the second row is the upper bound. Any values in df1 outside the lower and upper bound (column wise) needs to be replaced with NaN.
So in this example the output should be:
A B C
0 nan 2 nan
1 7 3 6
2 3 nan nan
As a basic I am trying df1[df1 < df2] = np.nan, but this does not work. I have also tried .where() but not getting any success.
Would appreciate some help here, thanks.
IIUC
df=df1.where(df1.ge(df2.iloc[0])&df1.lt(df2.iloc[1]))
A B C
0 NaN 2.0 NaN
1 7.0 3.0 6.0
2 3.0 NaN NaN
You could do something like:
lower = df1 < df2.iloc[0, :]
upper = df1 > df2.iloc[1, :]
df1[lower | upper] = np.nan
print(df1)
Output
A B C
0 NaN 2.0 NaN
1 7.0 3.0 6.0
2 3.0 NaN NaN
Here is one with df.clip and mask:
df1.mask(df1.ne(df1.clip(lower = df2.loc[0],upper = df1.loc[1],axis=1)))
A B C
0 NaN 2.0 NaN
1 7.0 3.0 6.0
2 3.0 NaN NaN
A slightly different approach using between,
df1.apply(lambda x:x.where(x.between(*df2.values, False)), axis=1)
I'm working with a dataset with ~80 columns, many of which contain NaN. I definitely don't want to manually inspect dtype for each column and impute based on that.
So I wrote a function to impute a column's missing values based on its dtype:
def impute_df(df, col):
# if col is float, impute mean
if df[col].dtype == "int64":
df[col].fillna(df[col].mean(), inplace=True)
else:
df[col].fillna(df[col].mode()[0], inplace=True)
But to use this, I'd have to loop over all columns in my DataFrame, something like:
for col in train_df.columns:
impute_df(train_df, col)
And I know looping in Pandas is generally slow. Is there a better way of going about this?
Thanks!
I think you need select_dtypes for numeric and non numeric columns and then apply fillna for filtered columns:
df = pd.DataFrame({'A':list('abcdef'),
'B':[np.nan,5,4,5,5,4],
'C':[7,8,np.nan,4,2,3],
'D':[1,3,5,7,1,0],
'E':[5,3,6,9,2,4],
'F':['a','a','b','b','b',np.nan]})
print (df)
A B C D E F
0 a NaN 7.0 1 5 a
1 b 5.0 8.0 3 3 a
2 c 4.0 NaN 5 6 b
3 d 5.0 4.0 7 9 b
4 e 5.0 2.0 1 2 b
5 f 4.0 3.0 0 4 NaN
cols1 = df.select_dtypes([np.number]).columns
cols2 = df.select_dtypes(exclude = [np.number]).columns
df[cols1] = df[cols1].fillna(df[cols1].mean())
df[cols2] = df[cols2].fillna(df[cols2].mode().iloc[0])
print (df)
A B C D E F
0 a 4.6 7.0 1 5 a
1 b 5.0 8.0 3 3 a
2 c 4.0 4.8 5 6 b
3 d 5.0 4.0 7 9 b
4 e 5.0 2.0 1 2 b
5 f 4.0 3.0 0 4 b
I think you do not need a function here,
for example:
df=pd.DataFrame({'A':[1,np.nan,3,4],'A_1':[1,np.nan,3,4],'B':['A','A',np.nan,'B']})
v=df.select_dtypes(exclude=['object']).columns
t=~df.columns.isin(v)
df.loc[:,v]=df.loc[:,v].fillna(df.loc[:,v].mean().to_dict())
df.loc[:,t]=df.loc[:,t].fillna(df.loc[:,t].mode().iloc[0].to_dict())
df
Out[1440]:
A A_1 B
0 1.000000 1.000000 A
1 2.666667 2.666667 A
2 3.000000 3.000000 A
3 4.000000 4.000000 B
Using 'bfill' or 'ffill' on a groupby element is trivial, but what if you need to fill the na with a specific value in a second column, based on a condition in a third column?
For example:
>>> df=pd.DataFrame({'date':['01/10/2017', '02/09/2017', '02/10/2016','01/10/2017', '01/11/2017', '02/10/2016'], 'a':[1,1,1,2,2,2], 'b':[4,np.nan,6, 5, np.nan, 7]})
>>> df
a b date
0 1 4.0 01/10/2017
1 1 NaN 02/09/2017
2 1 6.0 02/10/2016
3 2 5.0 01/10/2017
4 2 NaN 01/11/2017
5 2 7.0 02/10/2016
I need to group by column 'a', and fill the NaN with the column 'b' value where the date for that row is closest to the date in the NaN row.
So the output should look like:
a b date
0 1 4.0 01/10/2017
1 1 6.0 02/09/2017
2 1 6.0 02/10/2016
3 2 5.0 01/10/2017
4 2 5.0 01/11/2017
5 2 7.0 02/10/2016
Assume there is a closest_date() function that takes the NaN date and the list of other dates in that group, and returns the closest date.
I'm trying to find a clean solution that doesn't have to iterate through rows, ideally able to use apply() with lambdas. Any ideas?
This should work:
df['closest_date_by_a'] = df.groupby('a')['date'].apply(closest_date)
df['b'] = df.groupby(['a', 'closest_date_by_a'])['b'].ffill().bfill()
Given a function (closest_date()), you need to apply that function by group so it calculates the closest dates for rows within each group. Then you can group by both the main grouping column (a) and the closest date column (closest_date_by_a) and perform your filling.
Ensure that your date column are in fact dates.
df = pd.DataFrame(
{'date': ['01/10/2017', '02/09/2017', '02/10/2016','01/10/2017', '01/11/2017', '02/10/2016'],
'a':[1,1,1,2,2,2], 'b':[4,np.nan,6, 5, np.nan, 7]})
df.date = pd.to_datetime(df.date)
print(df)
a b date
0 1 4.0 2017-01-10
1 1 NaN 2017-02-09
2 1 6.0 2016-02-10
3 2 5.0 2017-01-10
4 2 NaN 2017-01-11
5 2 7.0 2016-02-10
Use reindex with method='nearest' after having dropna()
def fill_with_nearest(df):
s = df.set_index('date').b
s = s.dropna().reindex(s.index, method='nearest')
s.index = df.index
return s
df.loc[df.b.isnull(), 'b'] = df.groupby('a').apply(fill_with_nearest).reset_index(0, drop=True)
print(df)
a b date
0 1 4.0 2017-01-10
1 1 4.0 2017-02-09
2 1 6.0 2016-02-10
3 2 5.0 2017-01-10
4 2 5.0 2017-01-11
5 2 7.0 2016-02-10
I have these two dataframes:
import pandas as pd
df1 = pd.DataFrame({'A':[1,2,4,6]})
df2 = pd.DataFrame({'A':[1,2,3,6]})
df1
Out[27]:
A
0 1
1 2
2 4
3 6
df2
Out[28]:
A
0 1
1 2
2 3
3 6
I want to merge them in a way that both columns are preserved, common values are joined disregarding their index and unique values get preserved with a fill-value in the other row, that is, I want this result:
A_x A_y
0 1.0 1.0
1 2.0 2.0
2 NaN 3.0
3 4.0 NaN
4 6.0 6.0
I have tried
pd.merge(df1,df2,on=['A'],how='outer')
pd.concat([df1,df2],axis=1,join='outer')
but those two don't yield the desired result. I've tried them with different options but no luck.
I also looked into other methods like append and assign but none seems to provide the functionality to do this.
I feel like this is a common operation that should have an easy straightforward solution, so I might be overlooking something obvious.
Can you tell me how it's done right?
Solution with concat which concatenate values by index, so set_index is necessary:
df = pd.concat([df1.set_index('A', drop=False).A,
df2.set_index('A', drop=False).A],
axis=1,
keys=('A_x','A_y')).reset_index(drop=True)
print (df)
A_x A_y
0 1.0 1.0
1 2.0 2.0
2 NaN 3.0
3 4.0 NaN
4 6.0 6.0
df2 = df2.set_index('A', drop=False)
kws = dict(on='A', lsuffix='_x', rsuffix='_y', how='outer')
df1.join(df2, **kws).drop('A', 1)
A_x A_y
0 1.0 1.0
1 2.0 2.0
2 4.0 NaN
3 6.0 6.0
3 NaN 3.0