I need to regroup a df from the above format in the one below but it fails and the output shape is (unique number of IDs, 2). Is there a more obvious solution?
You can use groupby and pivot:
(df.assign(n=df.groupby('ID').cumcount().add(1))
.pivot(index='ID', columns='n', values='Value')
.add_prefix('val_')
.reset_index()
)
Example input:
df = pd.DataFrame({'ID': [7,7,8,11,12,18,22,22,22],
'Value': list('abcdefghi')})
Output:
n ID val_1 val_2 val_3
0 7 a b NaN
1 8 c NaN NaN
2 11 d NaN NaN
3 12 e NaN NaN
4 18 f NaN NaN
5 22 g h i
Related
I am new to Pandas and I am stuck at this specific problem where I have 2 DataFrames in Pandas, e.g.
>>> df1
A B
0 1 9
1 2 6
2 3 11
3 4 8
>>> df2
A B
0 Nan 0.05
1 Nan 0.05
2 0.16 Nan
3 0.16 Nan
What I am trying to achieve is to retain all values from df1 except where there is a NaN in df2 i.e.
>>> df3
A B
0 Nan 9
1 Nan 6
2 3 Nan
3 4 Nan
I am talking about dfs with 10,000 rows each so I can't do this manually. Also indices and columns are the exact same in each case. I also have no NaN values in df1.
As far as I understand df.update() will either overwrite all values including NaN or update only those that are NaN.
You can use boolean masking using DataFrame.notna.
# df2 = df2.astype(float) # This needed if your dtypes are not floats.
m = df2.notna()
df1[m]
A B
0 NaN 9.0
1 NaN 6.0
2 3.0 NaN
3 4.0 NaN
I need to change the following data frame in which one column contains a list of tuple
df = pd.DataFrame({'columns1':list('AB'),'columns2':[1,2],
'columns3':[[(122,0.5), (104, 0)], [(104, 0.6)]]})
print (df)
columns1 columns2 columns3
0 A 1 [(122, 0.5), (104, 0)]
1 B 2 [(104, 0.6)]
in to this, in which the tuple first element should be the column header
columns1 columns2 104 122
0 A 1 0.0 0.5
1 B 2 0.6 NaN
How can I do this using panda in Jupiter notebook
Use list comprehension with convert values to dictionaries, sorting columns and add to original with DataFrame.join:
df = pd.read_csv('Sample - Sample.csv.csv')
print (df)
column1 column2 column3
0 A U1 [(187, 0.674), (111, 0.738)]
1 B U2 [(54, 1.0)]
2 C U3 [(169, 0.474), (107, 0.424), (88, 0.519), (57,...
import ast
df1 = pd.DataFrame([dict(ast.literal_eval(x)) for x in df.pop('column3')], index=df.index).sort_index(axis=1)
df = df.join(df1)
print (df)
column1 column2 54 57 64 88 107 111 169 187
0 A U1 NaN NaN NaN NaN NaN 0.738 NaN 0.674
1 B U2 1.0 NaN NaN NaN NaN NaN NaN NaN
2 C U3 NaN 0.526 0.217 0.519 0.424 NaN 0.474 NaN
I have the following dataframe:
ID col_1
1 NaN
2 NaN
3 4.0
2 NaN
2 NaN
3 NaN
3 3.0
1 NaN
I need the following output:
ID col_1
1 NaN
1 NaN
2 NaN
2 NaN
2 NaN
how to do this in pandas
You can create a boolean mask with isna then group this mask by ID and transform using all, then you can filter the rows with the help of this mask:
mask = df['col_1'].isna().groupby(df['ID']).transform('all')
df[mask].sort_values('ID')
Alternatively you can use groupby + filter to filter out the groups which satisfy the condition where all values in col_1 are NaN but this method should be slower than the above:
df.groupby('ID').filter(lambda g: g['col_1'].isna().all()).sort_values('ID')
ID col_1
0 1 NaN
7 1 NaN
1 2 NaN
3 2 NaN
4 2 NaN
Let us try with isin after groupby with all
s = df['col_1'].isna().groupby(df['ID']).all()
df = df.loc[df.ID.isin(s[s].index.tolist())]
df
Out[73]:
ID col_1
0 1 NaN
1 2 NaN
3 2 NaN
4 2 NaN
7 1 NaN
import pandas as pd
import numpy as np
df=pd.read_excel(r"D:\Stack_overflow\test12.xlsx")
df1=(df[df['cols_1'].isnull()]).sort_values(by=['ID'])
I think we can simply take out the null values.
I want to clean some data by replacing only CONSECUTIVE 0s in a data frame
Given:
import pandas as pd
import numpy as np
d = [[1,np.NaN,3,4],[2,0,0,np.NaN],[3,np.NaN,0,0],[4,np.NaN,0,0]]
df = pd.DataFrame(d, columns=['a', 'b', 'c', 'd'])
df
a b c d
0 1 NaN 3 4.0
1 2 0.0 0 NaN
2 3 NaN 0 0.0
3 4 NaN 0 0.0
The desired result should be:
a b c d
0 1 NaN 3 4.0
1 2 0.0 NaN NaN
2 3 NaN NaN NaN
3 4 NaN NaN NaN
where column c & d are affected but column b is NOT affected as it only has 1 zero (and not consecutive 0s).
I have experimented with this answer:
Replacing more than n consecutive values in Pandas DataFrame column
which is along the right lines but the solution keeps the first 0 in a given column which is not desired in my case.
Let us do shift with mask
df=df.mask((df.shift().eq(df)|df.eq(df.shift(-1)))&(df==0))
Out[469]:
a b c d
0 1 NaN 3.0 4.0
1 2 0.0 NaN NaN
2 3 NaN NaN NaN
3 4 NaN NaN NaN
Trying to construct pandas DataFrame from list of dicts
List of dicts:
a = [{'1': 'A'},
{'2': 'B'},
{'3': 'C'}]
Pass list of dicts into pd.DataFrame():
df = pd.DataFrame(a)
Actual results:
1 2 3
0 A NaN NaN
1 NaN B NaN
2 NaN NaN C
pd.DataFrame(a, columns=['Key', 'Value'])
Actual results:
Key Value
0 NaN NaN
1 NaN NaN
2 NaN NaN
Expected results:
Key Value
0 1 A
1 2 B
2 3 C
try this,
from collections import ChainMap
data = dict(ChainMap(*a))
pd.DataFrame(data.items(), columns= ['Key','Value'])
O/P:
Key Value
0 1 A
1 2 B
2 3 C
Something like this with a list comprehension:
pd.DataFrame(([(x, y) for i in a for x, y in i.items()]),columns=['Key','Value'])
Key Value
0 1 A
1 2 B
2 3 C