I have a multiindex dataframe, an example can be created using:
arrays = [['bar', 'bar', 'bar', 'bar', 'bar','baz', 'baz','baz', 'baz', 'baz', 'foo', 'foo', 'foo',
'foo', 'foo', 'qux', 'qux', 'qux','qux', 'qux'],
[1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5]]
tuples = list(zip(*arrays))
values = [1,1,2,2,2,1,1,1,1,1,2,2,2,3,3,3,2,2,2,1]
df = pd.DataFrame(values, index=pd.MultiIndex.from_tuples(tuples, names=['first', 'second']),
columns = ['test'])
resulting in a dataframe that looks like this
test
first sec
bar 1 1
2 1
3 2
4 2
5 2
baz 1 1
2 1
3 1
4 1
5 1
foo 1 2
2 2
3 2
4 3
5 3
qux 1 3
2 2
3 2
4 2
5 2
I would like to figure out how to get the cumulative sum of the numbers in "test" for all "first" in a new column called ['result']. I feel like I am close using
df['result'] = df.test.expanding(1).sum()
but I cannot figure out how to cut it off at df['sec'] = 5 and start again (it just keeps going)
I would like my final output to look like
test result
first sec
bar 1 1 1
2 1 2
3 2 4
4 2 6
5 2 8
baz 1 1 1
2 1 2
3 1 3
4 1 4
5 1 5
foo 1 2 2
2 2 4
3 2 6
4 3 9
5 3 12
qux 1 3 3
2 2 5
3 2 7
4 2 9
5 2 11
Suggestions are appreciated.
Did this work,
df['result'] = df.groupby(['first'])['test'].transform(lambda x: x.cumsum())
Related
I would like to replace the values in my pd.DataFrame, df with counts of the value in row.
import pandas as pd
df = pd.DataFrame({
'foo': [3,3,1,1,1,2],
'bar': [4,4,1,3,3,3]
}).transpose()
0
1
2
3
4
5
foo
3
3
1
1
1
2
bar
4
4
1
3
3
3
I would expect to see:
0
1
2
3
4
5
foo
2
2
3
3
3
1
bar
2
2
1
3
3
3
Unable to determine a solution using .apply().
What is the most sensible way of achieving the above?
One way using pandas.Series.value_counts with map:
df.apply(lambda x: x.map(x.value_counts()), axis=1)
Output:
0 1 2 3 4 5
foo 2 2 3 3 3 1
bar 2 2 1 3 3 3
I have a df like this:
Index Parameters A B C D E
1 Apple 1 2 3 4 5
2 Banana 2 4 5 3 5
3 Potato 3 5 3 2 1
4 Tomato 1 1 1 1 1
5 Pear 4 5 5 4 3
I want to add all the rows which has Parameter values as "Apple" , "Banana" and "Pear".
Output:
Index Parameters A B C D E
1 Apple 1 2 3 4 5
2 Banana 2 4 5 3 5
3 Potato 3 5 3 2 1
4 Tomato 1 1 1 1 1
5 Pear 4 5 5 4 3
6 Total 7 11 13 11 13
My Effort:
df[:,'Total'] = df.sum(axis=1) -- Works but I want specific values only and not all
Tried by the index in my case 1,2 and 5 but in my original df the index can vary from time to time and hence rejected that solution.
Saw various answers on SO but none of them could solve my problem!!
First idea is create index by Parameters column and select rows for sum and last convert index to column:
L = ["Apple" , "Banana" , "Pear"]
df = df.set_index('Parameters')
df.loc['Total'] = df.loc[L].sum()
df = df.reset_index()
print (df)
Parameters A B C D E
0 Apple 1 2 3 4 5
1 Banana 2 4 5 3 5
2 Potato 3 5 3 2 1
3 Tomato 1 1 1 1 1
4 Pear 4 5 5 4 3
5 Total 7 11 13 11 13
Or add new row for filtered rows by membership with Series.isin and overwrite last added value by Total:
last = len(df)
df.loc[last] = df[df['Parameters'].isin(L)].sum()
df.loc[last, 'Parameters'] = 'Total'
print (df)
Parameters A B C D E
Index
1 Apple 1 2 3 4 5
2 Banana 2 4 5 3 5
3 Potato 3 5 3 2 1
4 Tomato 1 1 1 1 1
5 Total 7 11 13 11 13
Another similar solution is filtering all columns without first and add value in one element list:
df.loc[len(df)] = ['Total'] + df.iloc[df['Parameters'].isin(L).values, 1:].sum().tolist()
I have a dataframe looks like,
A B
1 2
1 3
1 4
2 5
2 6
3 7
3 8
If I df.groupby('A'), how do I turn each group into sub-dataframes, so it will look like, for A=1
A B
1 2
1 3
1 4
for A=2,
A B
2 5
2 6
for A=3,
A B
3 7
3 8
By using get_group
g=df.groupby('A')
g.get_group(1)
Out[367]:
A B
0 1 2
1 1 3
2 1 4
You are close, need convert groupby object to dictionary of DataFrames:
dfs = dict(tuple(df.groupby('A')))
print (dfs[1])
A B
0 1 2
1 1 3
2 1 4
print (dfs[2])
A B
3 2 5
4 2 6
Hi all I have the following dataframe:
A | B | C
1 2 3
2 3 4
3 4 5
4 5 6
And I am trying to only repeat the last two rows of the data so that it looks like this:
A | B | C
1 2 3
2 3 4
3 4 5
3 4 5
4 5 6
4 5 6
I have tried using append, concat and repeat to no avail.
repeated = lambda x:x.repeat(2)
df.append(df[-2:].apply(repeated),ignore_index=True)
This returns the following dataframe, which is incorrect:
A | B | C
1 2 3
2 3 4
3 4 5
4 5 6
3 4 5
3 4 5
4 5 6
4 5 6
You can use numpy.repeat for repeating index and then create df1 by loc, last append to original, but before filter out last 2 rows by iloc:
df1 = df.loc[np.repeat(df.index[-2:].values, 2)]
print (df1)
A B C
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
print (df.iloc[:-2])
A B C
0 1 2 3
1 2 3 4
df = df.iloc[:-2].append(df1,ignore_index=True)
print (df)
A B C
0 1 2 3
1 2 3 4
2 3 4 5
3 3 4 5
4 4 5 6
5 4 5 6
If want use your code add iloc for filtering only last 2 rows:
repeated = lambda x:x.repeat(2)
df = df.iloc[:-2].append(df.iloc[-2:].apply(repeated),ignore_index=True)
print (df)
A B C
0 1 2 3
1 2 3 4
2 3 4 5
3 3 4 5
4 4 5 6
5 4 5 6
Use pd.concat and index slicing with .iloc:
pd.concat([df,df.iloc[-2:]]).sort_values(by='A')
Output:
A B C
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
I'm partial to manipulating the index into the pattern we are aiming for then asking the dataframe to take the new form.
Option 1
Use pd.DataFrame.reindex
df.reindex(df.index[:-2].append(df.index[-2:].repeat(2)))
A B C
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
Same thing in multiple lines
i = df.index
idx = i[:-2].append(i[-2:].repeat(2))
df.reindex(idx)
Could also use loc
i = df.index
idx = i[:-2].append(i[-2:].repeat(2))
df.loc[idx]
Option 2
Reconstruct from values. Only do this is all dtypes are the same.
i = np.arange(len(df))
idx = np.append(i[:-2], i[-2:].repeat(2))
pd.DataFrame(df.values[idx], df.index[idx])
0 1 2
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
Option 3
Can also use np.array in iloc
i = np.arange(len(df))
idx = np.append(i[:-2], i[-2:].repeat(2))
df.iloc[idx]
A B C
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
Given the following data frame:
import pandas as pd
import numpy as np
df = pd.DataFrame({'A':[1,2,3],
'B':[4,5,6],
'C':[7,8,9],
'D':[1,3,5],
'E':[5,3,6],
'F':[7,4,3]})
df
A B C D E F
0 1 4 7 1 5 7
1 2 5 8 3 3 4
2 3 6 9 5 6 3
How can one assign column names to variables for use in referring to said column names?
For example, if I do this:
cols=['A','B']
cols2=['C','D']
I then want to do something like this:
df[cols,'F',cols2]
But the result is this:
TypeError: unhashable type: 'list'
I think you need add column F to list:
allcols = cols + ['F'] + cols2
print df[allcols]
A B F C D
0 1 4 7 7 1
1 2 5 4 8 3
2 3 6 3 9 5
Or:
print df[cols + ['F'] +cols2]
A B F C D
0 1 4 7 7 1
1 2 5 4 8 3
2 3 6 3 9 5
Need give a list with columns for reference.
In [48]: df[cols+['F']+cols2]
Out[48]:
A B F C D
0 1 4 7 7 1
1 2 5 4 8 3
2 3 6 3 9 5
and, consider using df.loc[:, cols+['F']+cols2], df.ix[:, cols+['F']+cols2] for slicing.
Python 3 solution:
In [154]: df[[*cols,'F',*cols2]]
Out[154]:
A B F C D
0 1 4 7 7 1
1 2 5 4 8 3
2 3 6 3 9 5