I have:
df = pd.DataFrame(
[
[22, 33, 44],
[55, 11, 22],
[33, 55, 11],
],
index=["abc", "def", "ghi"],
columns=list("abc")
) # size(3,3)
and:
unique = pd.Series([11, 22, 33, 44, 55]) # size(1,5)
then I create a new df based on unique and df, so that:
df_new = pd.DataFrame(index=unique, columns=df.columns) # size(5,3)
From this newly created df, I'd like to create a new boolean df based on unique and df, so that the end result is:
df_new = pd.DataFrame(
[
[0, 1, 1],
[1, 0, 1],
[1, 1, 0],
[0, 0, 1],
[1, 1, 0],
],
index=unique,
columns=df.columns
)
This new df is either true or false depending on whether the value is present in the original dataframe or not. For example, the first column has three values: [22, 55, 33]. In a df with dimensions (5,3), this first column would be: [0, 1, 1, 0, 1] i.e. [0, 22, 33, 0 , 55]
I tried filter2 = unique.isin(df) but this doesn't work, also notnull. I tried applying a filter but the dimensions returned were incorrect. How can I do this?
Use DataFrame.stack with DataFrame.reset_index, DataFrame.pivot, then check if not missing values by DataFrame.notna, cast to integers for True->1 and False->0 mapping and last remove index and columns names by DataFrame.rename_axis:
df_new = (df.stack()
.reset_index(name='v')
.pivot('v','level_1','level_0')
.notna()
.astype(int)
.rename_axis(index=None, columns=None))
print (df_new)
a b c
11 0 1 1
22 1 0 1
33 1 1 0
44 0 0 1
55 1 1 0
Helper Series is not necessary, but if there is more values or is necessary change order by helper Series use add DataFrame.reindex:
#added 66
unique = pd.Series([11, 22, 33, 44, 55,66])
df_new = (df.stack()
.reset_index(name='v')
.pivot('v','level_1','level_0')
.reindex(unique)
.notna()
.astype(int)
.rename_axis(index=None, columns=None))
print (df_new)
a b c
11 0 1 1
22 1 0 1
33 1 1 0
44 0 0 1
55 1 1 0
66 0 0 0
Related
I have a dictionary with 2 DF : "quantity variation in %" and "prices". They are both symmetrical DF.
Let's say I want to set the price = 0 if the quantity variation in percentage is greater than 100 %
import numpy as np; import pandas as pd
d = {'qty_pct': pd.DataFrame({ '2020': [200, 0.5, 0.4],
'2021': [0.9, 0.5, 500],
'2022': [0.9, 300, 0.4]}),
'price': pd.DataFrame({ '2020': [-6, -2, -9],
'2021': [ 2, 3, 4],
'2022': [ 4, 6, 8]})}
# I had something like that in mind ...
df = d['price'].applymap(lambda x: 0 if x[d['qty_pct']] >=1 else x)
P.S. If by any chance there is a way to do this on asymmetrical DF, I would be curious to see how it's done.
Thanks,
I want to obtain this DF :
price = pd.DataFrame({'2020': [ 0, -2, -9],
'2021': [ 2, 3, 0],
'2022': [ 4, 0, 8]})
Assume price and qty_pct always have the same dimension, then you can just do:
d['price'][d['qty_pct'] >= 1] = 0
d['price']
2020 2021 2022
0 0 2 4
1 -2 3 0
2 -9 0 8
As an example, how do I convert df to df1, by gathering rows into matrices based on shared values in a specific column tidx?
>>> df = pd.DataFrame({'col3':[[1,40],[2,50],[3,60],[4,70]], 'tidx':[21,22,23,21]})
>>> df['col3'] = df['col3'].apply(np.array)
>>> df
col3 tidx
0 [1, 40] 21
1 [2, 50] 22
2 [3, 60] 23
3 [4, 70] 21
>>> df1 = pd.DataFrame({'col3':[[[1,40],[4,70]],[[2,50]],[[3,60]]], 'tidx':[21,22,23]})
>>> df1['col3'] = df1['col3'].apply(np.array)
>>> df1
col3 tidx
0 [[1, 40], [4, 70]] 21
1 [[2, 50]] 22
2 [[3, 60]] 23
You can use .groupby and then apply list function as shown in example below.
df = pd.DataFrame({'col3':[[1,40],[2,50],[3,60],[4,70]], 'tidx':[21,22,23,21]})
df1 = df.groupby('tidx')['col3'].apply(list).reset_index()
This is a similar question to this thread.
Lets consider df as:
df = pd.DataFrame([["a", 2, 3], ["b", 5, 6], ["c", 8, 9],["a", 0, 0], ["a", 8, 7], ["c", 2, 1]], columns = ["A", "B", "C"])
How can you calculate the difference between all rows and the row at Nth index in a group (lowest index for EACH group) for column "B", and put it in column "D"? I want to calculate mean square displacement for my data and I want to calculate the difference of values in a column in each group with the first appeared row in that group.
I tried:
df['D'] = df.groupby(["A"])['B'].sub(df.groupby(['A'])["B"].iloc[0])
Group = df.groupby(["A"])
However using .sub and groupby raise the following error:
AttributeError: 'SeriesGroupBy' object has no attribute 'sub'
the desired result would be like this:
A B C D
0 a 2 3 0 *lowest index in group "a"
1 b 5 6 0 *lowest index in group "b"
2 c 8 9 0 *lowest index in group "c"
3 a 0 0 -2
4 a 8 7 6
5 c 2 1 -6
I guess this answer could be enough of a hint for you:
import pandas as pd
df = pd.DataFrame([["a", 2, 3], ["b", 5, 6], ["c", 8, 9],["a", 0, 0], ["a", 8, 7], ["c", 2, 1]], columns = ["A", "B", "C"])
print("df:")
print(df)
print()
groupA = df.groupby(['A'])
print("groupA:")
print(groupA.groups)
print()
print("lowest indices for each group from columnA:")
lowest_indices = dict()
for k, v in groupA.groups.items():
lowest_indices[k] = v[0]
print(lowest_indices)
print()
columnB = df['B']
print("columnB:")
print(columnB)
print()
df['D'] = df['B']
for i in range(len(df)):
group_at_i = df['A'].iloc[i]
lowest_index_of_that = lowest_indices[group_at_i]
b_element_at_that_index = df['B'].iloc[lowest_index_of_that]
the_difference = df['B'].iloc[i] - b_element_at_that_index
df.loc[i, 'D'] = the_difference
print("df:")
print(df)
I have a following problem. I have this df:
d = {'id': [1, 1, 2, 2, 3], 'value': [0, 1, 0, 0, 1]}
df = pd.DataFrame(data=d)
I would like to have a new column where value will be 1 if in any other cases it is also 1. See desired output:
d = {'id': [1, 1, 2, 2, 3], 'value': [0, 1, 0, 0, 1], 'newvalue': [1, 1, 0, 0, 1]}
df = pd.DataFrame(data=d)
How can I do it please?
If need set 0,1 by condition - here at least one 1 use GroupBy.transform with GroupBy.any for mask and casting to integers for True, False to 1,0 map:
df['newvalue'] = df['value'].eq(1).groupby(df['id']).transform('any').astype(int)
Alternative:
df['newvalue'] = df['id'].isin(df.loc[df['value'].eq(1), 'id']).astype(int)
Or if only 0,1 values is possible simplify solution for new column by maximal values per groups:
df['newvalue'] = df.groupby('id')['value'].transform('max')
print (df)
id value newvalue
0 1 0 1
1 1 1 1
2 2 0 0
3 2 0 0
4 3 1 1
I have the following dataframe with timeseries data:
df = pd.DataFrame(columns = ['id', 'value'])
df['value'] =[9, 16, 10, 12, 11, 14]
df['id'] = [1, 1, 1, 2, 2, 2]
For each timeseries (defined by column 'id' I want to calculate the variance to find timeseries that do not change at all or only very little.
The final dataframe should look like this:
df_end = pd.DataFrame(columns = ['id','value', 'var'])
df_end['value'] =[9, 16, 10, 12, 11, 14]
df_end['id'] = [1, 1, 1, 2, 2, 2]
df_end['var'] = [21, 21, 21, 2.3, 2.3, 2.3]
I tried:
df.groupby(df['id']).var()
which gives me the values, but I couldn't put it into the df in the right form. I am sure, there is a handy function for this that I don't know about yet!
Thanks for helping out!
Use GroupBy.transform with specify column value:
df['var'] = df.groupby('id')['value'].transform('var')
print (df)
id value var
0 1 9 14.333333
1 1 16 14.333333
2 1 10 14.333333
3 2 12 2.333333
4 2 11 2.333333
5 2 14 2.333333