Merge almost identical rows after removing nan values - python

I have a dataframe like,
pri_col col1 col2 Date
r1 3 4 2020-09-10
r1 4 1 2020-09-11
r1 2 7 2020-09-12
r1 6 4 2020-09-13
Note: There are many more unique values in 'pri_col' column. This is just a sample here. So I'm giving single value. Also, for a single value of 'pri_col' the value of 'Date' will be unique always.
I need the dataframe like,
pri_col col1_2020-09-10 col1_2020-09-11 col1_2020-09-12 col1_2020-09-13 col2_2020-09-10 col2_2020-09-11 col2_2020-09-12 col2_2020-09-13
r1 3 4 2 6 4 1 7 4
According to a previous solution, I have tried this solution:
df = (df.reset_index()
.melt(id_vars=['index','pri_col','Date'],
var_name='cols',
value_name='val')
.pivot(index=['index','pri_col'],
columns=['cols','Date'],
values='val'))
df.columns = [f'{a}_{b}' for a, b in df.columns]
df = df.reset_index(level=1).rename_axis(None)
print (df)
But this is the resulting dataframe:
pri_col col1_2020-09-10 col1_2020-09-11 col1_2020-09-12 col1_2020-09-13 col2_2020-09-10 col2_2020-09-11 col2_2020-09-12 col2_2020-09-13
r1 3 NaN NaN NaN 4 NaN NaN NaN
r1 NaN 4 NaN NaN NaN 1 NaN NaN
r1 NaN NaN 2 NaN NaN NaN 7 NaN
r1 NaN NaN NaN 6 NaN NaN NaN 4
How do I solve the issue?
Also, I had asked a question recently that may sound similar.


IIUC, use pandas.DataFrame.set_index with unstack:
new_df = df.set_index(['pri_col', 'Date']).unstack()
new_df.columns = ["%s_%s" % (i, j) for i, j in new_df.columns]
print(new_df)
Output:
col1_2020-09-10 col1_2020-09-11 col1_2020-09-12 col1_2020-09-13 \
pri_col
r1 3 4 2 6
col2_2020-09-10 col2_2020-09-11 col2_2020-09-12 col2_2020-09-13
pri_col
r1 4 1 7 4

Related

Create dataframe with hierarchical indices and extra columns from non-hierarchically indexed dataframe

Consider a simple dataframe:
import numpy as np
import pandas as pd
x = pd.DataFrame(np.arange(10).reshape(5,2))
print(x)
0 1
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9
I would like to create a hierarchically indexed dataframe of the form:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN
where the 'a' columns correspond to the original dataframe columns and the 'b' columns are blank (or nan).
I can certainly create a hierarchically indexed dataframe with all NaNs and loop over the columns of the original dataframe, writing them into
the new dataframe. Is there something more compact than that?

you can do with MultiIndex.from_product
extra_level = ['a', 'b']
new_cols = pd.MultiIndex.from_product([x.columns, extra_level])
x.columns = new_cols[::len(x.columns)] # select all the first element of extra_level
x = x.reindex(columns=new_cols)
print(x)
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN

Very much like #Ben.T I am using MultiIndex.from_product:
x.assign(l='a')
.set_index('l', append=True)
.unstack()
.reindex(pd.MultiIndex.from_product([x.columns.tolist(), ['a','b']]), axis=1)
Output:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN

Select rows with specific values in columns and include rows with NaN in pandas dataframe

I have a DataFrame df that looks something like this:
df
a b c
0 0.557894 -0.196294 -0.020490
1 1.138774 -0.699224 NaN
2 NaN 2.384483 0.554292
3 -0.069319 NaN 1.162941
4 1.040089 -0.271777 NaN
5 -0.337374 NaN -0.771888
6 -1.813278 -1.564666 NaN
7 NaN NaN NaN
8 0.737413 NaN 0.679575
9 -2.345448 2.443669 -1.409422
I want to select the rows that have a value over some value, which I would normally do using:
new_df = df[df['c'] >= .5]
but that will return:
a b c
2 NaN 2.384483 0.554292
3 -0.069319 NaN 1.162941
5 -0.337374 NaN 0.771888
8 0.737413 NaN 0.679575
I want to get those rows, but also keep the rows that have nan values in column 'c'. I haven't been able to find a question asking the same thing, they usually ask for one or the other, but not both. I can hard code the rows that I want to drop since I know the specific values, but I was wondering if there is a better solution. The end result should look something like this:
a b c
1 1.138774 -0.699224 NaN
2 NaN 2.384483 0.554292
3 -0.069319 NaN 1.162941
4 1.040089 -0.271777 NaN
6 -1.813278 -1.564666 NaN
7 NaN NaN NaN
8 0.737413 NaN 0.679575
Only dropping rows 0,5 and 9 since they are less than .5 in columns 'c'

You should use the | (or) operator.
import pandas as pd
import numpy as np
df = pd.DataFrame({'a': [0.557894,1.138774,np.nan,-0.069319,1.040089,-0.337374,-1.813278,np.nan,0.737413,-2.345448],
'b': [-0.196294,-0.699224,2.384483,np.nan,-0.271777,np.nan,-1.564666,np.nan,np.nan,2.443669],
'c': [-0.020490,np.nan,0.554292,1.162941,np.nan,-0.771888,np.nan,np.nan,0.679575,-1.409422]})
df = df[(df['c'] >= .5) | (df['c'].isnull())]
print(df)
Output:
a b c
1 1.138774 -0.699224 NaN
2 NaN 2.384483 0.554292
3 -0.069319 NaN 1.162941
4 1.040089 -0.271777 NaN
6 -1.813278 -1.564666 NaN
7 NaN NaN NaN
8 0.737413 NaN 0.679575

You should be able to do this by
new_df = df[df['c'] >=5 or df['c'] == 'NaN']

How to fill and merge df with 10 empty rows?

how to fill df with empty rows or create a df with empty rows.
have df :
df = pd.DataFrame(columns=["naming","type"])
how to fill this df with empty rows

Specify index values:
df = pd.DataFrame(columns=["naming","type"], index=range(10))
print (df)
naming type
0 NaN NaN
1 NaN NaN
2 NaN NaN
3 NaN NaN
4 NaN NaN
5 NaN NaN
6 NaN NaN
7 NaN NaN
8 NaN NaN
9 NaN NaN
If need empty strings:
df = pd.DataFrame('',columns=["naming","type"], index=range(10))
print (df)
naming type
0
1
2
3
4
5
6
7
8
9

Pandas: replace column A with column B if B is not missing

I have question similar to a previous post. I want to replace missing values in A with B if B is not-missing. I've used a toy dataset.
#Create sample dataset
import pandas as pd
import numpy as np
np.random.seed(12345)
df = pd.DataFrame(np.random.randn(100, 4), columns=list('ABCD'))
df
df[df < 0] = 'NaN'
print(df)
Obs. A B
0 NaN 0.478943
1 NaN NaN
2 1.96578 1.39341
3 0.0929079 0.281746
4 0.769023 1.24643
5 1.00719 NaN
6 0.274992 0.228913
7 1.35292 0.886429
8 NaN NaN
9 1.66903 NaN
#Replace NaN in A with B if B is not NaN
df['A'] = np.where(pd.isnull(df['A']) & pd.notnull(df['B']) == 0, df['B']*1, df['A'])
print(df)
obs A B
0 0.478943 0.478943
1 NaN NaN
2 1.39341 1.39341
3 0.281746 0.281746
4 1.24643 1.24643
5 NaN NaN
6 0.228913 0.228913
7 0.886429 0.886429
8 NaN NaN
9 NaN NaN
This code does the job. But why do I need pd.notnull(df['B']) == 0? If I write:
pd.notnull(df['B'])
instead, the code does not work correctly. The output from that is:
Obs. A B
0 NaN 0.478943
1 NaN NaN
2 1.96578 1.39341
3 0.0929079 0.281746
4 0.769023 1.24643
5 1.00719 NaN
6 0.274992 0.228913
7 1.35292 0.886429
8 NaN NaN
9 1.66903 NaN
I'm trying to understand the flaw in my logic. Any other simple intuitive code will be appreciated.
I basically need to do this simple operation for a very large dataset (100m obs+) so looking for a fast way (in terms of computer processing time) to do it. Thanks in advance.

Replace 'NaN' with np.nan and apply fillna on column A using column B
df = df.replace('NaN', np.nan)
df.A.fillna(df.B, inplace=True)
Output:
A B
0 0.478943 0.478943
1 NaN NaN
2 1.965781 1.393406
3 0.092908 0.281746
4 0.769023 1.246435
5 1.007189 NaN
6 0.274992 0.228913
7 1.352917 0.886429
8 NaN NaN
9 1.669025 NaN

Unmelt Pandas DataFrame

I have a pandas dataframe with two id variables:
df = pd.DataFrame({'id': [1,1,1,2,2,3],
'num': [10,10,12,13,14,15],
'q': ['a', 'b', 'd', 'a', 'b', 'z'],
'v': [2,4,6,8,10,12]})
id num q v
0 1 10 a 2
1 1 10 b 4
2 1 12 d 6
3 2 13 a 8
4 2 14 b 10
5 3 15 z 12
I can pivot the table with:
df.pivot('id','q','v')
And end up with something close:
q a b d z
id
1 2 4 6 NaN
2 8 10 NaN NaN
3 NaN NaN NaN 12
However, what I really want is (the original unmelted form):
id num a b d z
1 10 2 4 NaN NaN
1 12 NaN NaN 6 NaN
2 13 8 NaN NaN NaN
2 14 NaN 10 NaN NaN
3 15 NaN NaN NaN 12
In other words:
'id' and 'num' my indices (normally, I've only seen either 'id' or 'num' being the index but I need both since I'm trying to retrieve the original unmelted form)
'q' are my columns
'v' are my values in the table
Update
I found a close solution from Wes McKinney's blog:
df.pivot_table(index=['id','num'], columns='q')
v
q a b d z
id num
1 10 2 4 NaN NaN
12 NaN NaN 6 NaN
2 13 8 NaN NaN NaN
14 NaN 10 NaN NaN
3 15 NaN NaN NaN 12
However, the format is not quite the same as what I want above.

You could use set_index and unstack
In [18]: df.set_index(['id', 'num', 'q'])['v'].unstack().reset_index()
Out[18]:
q id num a b d z
0 1 10 2.0 4.0 NaN NaN
1 1 12 NaN NaN 6.0 NaN
2 2 13 8.0 NaN NaN NaN
3 2 14 NaN 10.0 NaN NaN
4 3 15 NaN NaN NaN 12.0

You're really close slaw. Just rename your column index to None and you've got what you want.
df2 = df.pivot_table(index=['id','num'], columns='q')
df2.columns = df2.columns.droplevel().rename(None)
df2.reset_index().fillna("null").to_csv("test.csv", sep="\t", index=None)
Note that the the 'v' column is expected to be numeric by default so that it can be aggregated. Otherwise, Pandas will error out with:
DataError: No numeric types to aggregate
To resolve this, you can specify your own aggregation function by using a custom lambda function:
df2 = df.pivot_table(index=['id','num'], columns='q', aggfunc= lambda x: x)

you can remove name q.
df1.columns=df1.columns.tolist()
Zero's answer + remove q =
df1 = df.set_index(['id', 'num', 'q'])['v'].unstack().reset_index()
df1.columns=df1.columns.tolist()
id num a b d z
0 1 10 2.0 4.0 NaN NaN
1 1 12 NaN NaN 6.0 NaN
2 2 13 8.0 NaN NaN NaN
3 2 14 NaN 10.0 NaN NaN
4 3 15 NaN NaN NaN 12.0

This might work just fine:
Pivot
df2 = (df.pivot_table(index=['id', 'num'], columns='q', values='v')).reset_index())
Concatinate the 1st level column names with the 2nd
df2.columns =[s1 + str(s2) for (s1,s2) in df2.columns.tolist()]

Came up with a close solution
df2 = df.pivot_table(index=['id','num'], columns='q')
df2.columns = df2.columns.droplevel()
df2.reset_index().fillna("null").to_csv("test.csv", sep="\t", index=None)
Still can't figure out how to drop 'q' from the dataframe

It can be done in three steps:
#1: Prepare auxilary column 'id_num':
df['id_num'] = df[['id', 'num']].apply(tuple, axis=1)
df = df.drop(columns=['id', 'num'])
#2: 'pivot' is almost an inverse of melt:
df, df.columns.name = df.pivot(index='id_num', columns='q', values='v').reset_index(), ''
#3: Bring back 'id' and 'num' columns:
df['id'], df['num'] = zip(*df['id_num'])
df = df.drop(columns=['id_num'])
This is a result, but with different order of columns:
a b d z id num
0 2.0 4.0 NaN NaN 1 10
1 NaN NaN 6.0 NaN 1 12
2 8.0 NaN NaN NaN 2 13
3 NaN 10.0 NaN NaN 2 14
4 NaN NaN NaN 12.0 3 15
Alternatively with proper order:
def multiindex_pivot(df, columns=None, values=None):
#inspired by: https://github.com/pandas-dev/pandas/issues/23955
names = list(df.index.names)
df = df.reset_index()
list_index = df[names].values
tuples_index = [tuple(i) for i in list_index] # hashable
df = df.assign(tuples_index=tuples_index)
df = df.pivot(index="tuples_index", columns=columns, values=values)
tuples_index = df.index # reduced
index = pd.MultiIndex.from_tuples(tuples_index, names=names)
df.index = index
df = df.reset_index() #me
df.columns.name = '' #me
return df
df = df.set_index(['id', 'num'])
df = multiindex_pivot(df, columns='q', values='v')

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