I have this data:
import pandas as pd
import numpy as np
index = pd.MultiIndex.from_tuples(list(zip(*[['one', 'one', 'two', 'two'],['foo', 'bar', 'foo', 'bar']])))
df = pd.DataFrame(np.arange(12).reshape((3,4)), columns=index)
one two
foo bar foo bar
0 0 1 2 3
1 4 5 6 7
2 8 9 10 11
Is there a way to do simple vectorized calculations (like addition) for each level 0 group columns on each of the level 1 columns without having to reference the specific column level pairs like:
df[('one','add')] = df[('one','foo')]+df[('one','bar')]
I'd like to get
one two
foo bar add foo bar add
0 0 1 1 2 3 5
1 4 5 9 6 7 13
2 8 9 17 10 11 21
I fiddled around with it for a bit and here is a one-liner that solves the problem in my opinion. It's fully vectorized and doesn't address specific column names. It also puts the add column in the right place.
df.stack(0).assign(add=df.stack(0).sum(axis=1)).stack(0).unstack(0).T
Unfortunately, because of the property of stack / unstack to do the stacking / unstacking into the innermost level, it needs the cryptic .stack(0).unstack(0) operation. It seems like those two operations should cancel each other out, but they actually shuffle the index levels while preserving order.
Here is the same thing split into 3 lines without assign statement.
df = df.stack(0)
df['add'] = df.sum(axis=1)
df = df.stack(0).unstack(0).T
Use pandas.DataFrame.sum with axis=1 and level=0:
df2 = df.sum(axis=1, level=0)
print(df2)
Output:
one two
0 1 5
1 9 13
2 17 21
You can then add new column names to pandas.concat:
df2.columns = [(c, "add") for c in df2]
df2 = pd.concat([df, df2], 1).sort_index(1)
print(df2)
Output:
one two
add bar foo add bar foo
0 1 1 0 5 3 2
1 9 5 4 13 7 6
2 17 9 8 21 11 10
An alternative solution, here, using the same sum solution, but without pd.concat :
df[("one", "add")] = None
df[("two", "add")] = None
df.iloc[:, -2:] = df.sum(axis=1, level=0).to_numpy()
df.sort_index(1)
one two
add bar foo add bar foo
0 1.0 1 0 5.0 3 2
1 9.0 5 4 13.0 7 6
2 17.0 9 8 21.0 11 10
Related
I want to stack two columns on top of each other
So I have Left and Right values in one column each, and want to combine them into a single one. How do I do this in Python?
I'm working with Pandas Dataframes.
Basically from this
Left Right
0 20 25
1 15 18
2 10 35
3 0 5
To this:
New Name
0 20
1 15
2 10
3 0
4 25
5 18
6 35
7 5
It doesn't matter how they are combined as I will plot it anyway, and the new column name also doesn't matter because I can rename it.
You can create a list of the cols, and call squeeze to anonymise the data so it doesn't try to align on columns, and then call concat on this list, passing ignore_index=True creates a new index, otherwise you'll get the names as index values repeated:
cols = [df[col].squeeze() for col in df]
pd.concat(cols, ignore_index=True)
Many options, stack, melt, concat, ...
Here's one:
>>> df.melt(value_name='New Name').drop('variable', 1)
New Name
0 20
1 15
2 10
3 0
4 25
5 18
6 35
7 5
You can also use np.ravel:
import numpy as np
out = pd.DataFrame(np.ravel(df.values.T), columns=['New name'])
print(out)
# Output
New name
0 20
1 15
2 10
3 0
4 25
5 18
6 35
7 5
Update
If you have only 2 cols:
out = pd.concat([df['Left'], df['Right']], ignore_index=True).to_frame('New name')
print(out)
# Output
New name
0 20
1 15
2 10
3 0
4 25
5 18
6 35
7 5
Solution with unstack
df2 = df.unstack()
# recreate index
df2.index = np.arange(len(df2))
A solution with masking.
# Your data
import numpy as np
import pandas as pd
df = pd.DataFrame({"Left":[20,15,10,0], "Right":[25,18,35,5]})
# Masking columns to ravel
df2 = pd.DataFrame({"New Name":np.ravel(df[["Left","Right"]])})
df2
New Name
0 20
1 25
2 15
3 18
4 10
5 35
6 0
7 5
I ended up using this solution, seems to work fine
df1 = dfTest[['Left']].copy()
df2 = dfTest[['Right']].copy()
df2.columns=['Left']
df3 = pd.concat([df1, df2],ignore_index=True)
How I can retrieve column names from a call to DataFrame apply without knowing them in advance?
What I'm trying to do is apply a mapping from column names to functions to arbitrary DataFrames. Those functions might return multiple columns. I would like to end up with a DataFrame that contains the original columns as well as the new ones, the amount and names of which I don't know at build-time.
Other solutions here are Series-based. I'd like to do the whole frame at once, if possible.
What am I missing here? Are the columns coming back from apply lost in destructuring unless I know their names? It looks like assign might be useful, but will likely require a lot of boilerplate.
import pandas as pd
def fxn(col):
return pd.Series(col * 2, name=col.name+'2')
df = pd.DataFrame({'A': range(0, 10), 'B': range(10, 0, -1)})
print(df)
# [Edit:]
# A B
# 0 0 10
# 1 1 9
# 2 2 8
# 3 3 7
# 4 4 6
# 5 5 5
# 6 6 4
# 7 7 3
# 8 8 2
# 9 9 1
df = df.apply(fxn)
print(df)
# [Edit:]
# Observed: columns changed in-place.
# A B
# 0 0 20
# 1 2 18
# 2 4 16
# 3 6 14
# 4 8 12
# 5 10 10
# 6 12 8
# 7 14 6
# 8 16 4
# 9 18 2
df[['A2', 'B2']] = df.apply(fxn)
print(df)
# [Edit: I am doubling column values, so missing something, but the question about the column counts stands.]
# Expected: new columns added. How can I do this at runtime without knowing column names?
# A B A2 B2
# 0 0 40 0 80
# 1 4 36 8 72
# 2 8 32 16 64
# 3 12 28 24 56
# 4 16 24 32 48
# 5 20 20 40 40
# 6 24 16 48 32
# 7 28 12 56 24
# 8 32 8 64 16
# 9 36 4 72 8
You need to concat the result of your function with the original df.
Use pd.concat:
In [8]: x = df.apply(fxn) # Apply function on df and store result separately
In [10]: df = pd.concat([df, x], axis=1) # Concat with original df to get all columns
Rename duplicate column names by adding suffixes:
In [82]: from collections import Counter
In [38]: mylist = df.columns.tolist()
In [41]: d = {a:list(range(1, b+1)) if b>1 else '' for a,b in Counter(mylist).items()}
In [62]: df.columns = [i+str(d[i].pop(0)) if len(d[i]) else i for i in mylist]
In [63]: df
Out[63]:
A1 B1 A2 B2
0 0 10 0 20
1 1 9 2 18
2 2 8 4 16
3 3 7 6 14
4 4 6 8 12
5 5 5 10 10
6 6 4 12 8
7 7 3 14 6
8 8 2 16 4
9 9 1 18 2
You can assign directly with:
df[df.columns + '2'] = df.apply(fxn)
Outut:
A B A2 B2
0 0 10 0 20
1 1 9 2 18
2 2 8 4 16
3 3 7 6 14
4 4 6 8 12
5 5 5 10 10
6 6 4 12 8
7 7 3 14 6
8 8 2 16 4
9 9 1 18 2
Alternatively, you can leverage the #MayankPorwal answer by using .add_suffix('2') to the output from your apply function:
pd.concat([df, df.apply(fxn).add_suffix('2')], axis=1)
which will return the same output.
In your function, name=col.name+'2' is doing nothing (it's basically returning just col * 2). That's because apply returns the values back to the original column.
Anyways, it's possible to take the MayankPorwal approach: pd.concat + managing duplicated columns (make them unique). Another possible way to do that:
# Use pd.concat as mentioned in the first answer from Mayank Porwal
df = pd.concat([df, df.apply(fxn)], axis=1)
# Rename duplicated columns
suffix = (pd.Series(df.columns).groupby(df.columns).cumcount()+1).astype(str)
df.columns = df.columns + suffix.rename('1', '')
which returns the same output, and additionally manage further duplicated columns.
Answer on the behalf of OP:
This code does what I wanted:
import pandas as pd
# Simulated business logic: for an input row, return a number of columns
# related to the input, and generate names for them, such that we don't
# know the shape of the output or the names of its columns before the call.
def fxn(row):
length = row[0]
indicies = [row.index[0] + str(i) for i in range(0, length)]
series = pd.Series([i for i in range(0, length)], index=indicies)
return series
# Sample data: 0 to 18, inclusive, counting by 2.
df1 = pd.DataFrame(list(range(0, 20, 2)), columns=['A'])
# Randomize the rows to simulate different input shapes.
df1 = df1.sample(frac=1)
# Apply fxn to rows to get new columns (with expand). Concat to keep inputs.
df1 = pd.concat([df1, df1.apply(fxn, axis=1, result_type='expand')], axis=1)
print(df1)
Currently have a dataframe that is countries by series, with values ranging from 0-25
I want to sort the df so that the highest values appear in the top left (first), while the lowest appear in the bottom right (last).
FROM
A B C D ...
USA 4 0 10 16
CHN 2 3 13 22
UK 2 1 8 14
...
TO
D C A B ...
CHN 22 13 2 3
USA 16 10 4 0
UK 14 8 2 1
...
In this, the column with the highest values is now first, and the same is true with the index.
I have considered reindexing, but this loses the 'Countries' Index.
D C A B ...
0 22 13 2 3
1 16 10 4 0
2 14 8 2 1
...
I have thought about creating a new column and row that has the Mean or Sum of values for that respective column/row, but is this the most efficient way?
How would I then sort the DF after I have the new rows/columns??
Is there a way to reindex using...
df_mv.reindex(df_mv.mean(or sum)().sort_values(ascending = False).index, axis=1)
... that would allow me to keep the country index, and simply sort it accordingly?
Thanks for any and all advice or assistance.
EDIT
Intended result organizes columns AND rows from largest to smallest.
Regarding the first row of the A and B columns in the intended output, these are supposed to be 2, 3 respectively. This is because the intended result interprets the A column as greater than the B column in both sum and mean (even though either sum or mean can be considered for the 'value' of a row/column).
By saying the higher numbers would be in the top left, while the lower ones would be in the bottom right, I simply meant this as a general trend for the resulting df. It is the columns and rows as whole however, that are the intended focus. I apologize for the confusion.
You could use:
rows_index=df.max(axis=1).sort_values(ascending=False).index
col_index=df.max().sort_values(ascending=False).index
new_df=df.loc[rows_index,col_index]
print(new_df)
D C A B
CHN 22 13 2 3
USA 16 10 4 0
UK 14 8 2 1
Use .T to transpose rows to columns and vice versa:
df = df.sort_values(df.max().idxmax(), ascending=False)
df = df.T
df = df.sort_values(df.columns[0], ascending=False).T
Result:
>>> df
D C B A
CHN 22 13 3 2
USA 16 10 0 4
UK 14 8 1 2
Here's another way, this time without transposing but using axis=1 as an argument:
df = df.sort_values(df.max().idxmax(), ascending=False)
df = df.sort_values(df.index[0], axis=1, ascending=False)
Using numpy:
arr = df.to_numpy()
arr = arr[np.max(arr, axis=1).argsort()[::-1], :]
arr = np.sort(arr, axis=1)[:, ::-1]
df1 = pd.DataFrame(arr, index=df.index, columns=df.columns)
print(df1)
Output:
A B C D
USA 22 13 3 2
CHN 16 10 4 0
UK 14 8 2 1
For a DataFrame in Pandas, how can I select both the first 5 values and last 5 values?
For example
In [11]: df
Out[11]:
A B C
2012-11-29 0 0 0
2012-11-30 1 1 1
2012-12-01 2 2 2
2012-12-02 3 3 3
2012-12-03 4 4 4
2012-12-04 5 5 5
2012-12-05 6 6 6
2012-12-06 7 7 7
2012-12-07 8 8 8
2012-12-08 9 9 9
How to show the first two and the last two rows?
You can use iloc with numpy.r_:
print (np.r_[0:2, -2:0])
[ 0 1 -2 -1]
df = df.iloc[np.r_[0:2, -2:0]]
print (df)
A B C
2012-11-29 0 0 0
2012-11-30 1 1 1
2012-12-07 8 8 8
2012-12-08 9 9 9
df = df.iloc[np.r_[0:4, -4:0]]
print (df)
A B C
2012-11-29 0 0 0
2012-11-30 1 1 1
2012-12-01 2 2 2
2012-12-02 3 3 3
2012-12-05 6 6 6
2012-12-06 7 7 7
2012-12-07 8 8 8
2012-12-08 9 9 9
You can use df.head(5) and df.tail(5) to get first five and last five.
Optionally you can create new data frame and append() head and tail:
new_df = df.tail(5)
new_df = new_df.append(df.head(5))
Not quite the same question but if you just want to show the top / bottom 5 rows (eg with display in jupyter or regular print, there's potentially a simpler way than this if you use the pd.option_context context.
#make 100 3d random numbers
df = pd.DataFrame(np.random.randn(100,3))
# sort them by their axis sum
df = df.loc[df.sum(axis=1).index]
with pd.option_context('display.max_rows',10):
print(df)
Outputs:
0 1 2
0 -0.649105 -0.413335 0.374872
1 3.390490 0.552708 -1.723864
2 -0.781308 -0.277342 -0.903127
3 0.433665 -1.125215 -0.290228
4 -2.028750 -0.083870 -0.094274
.. ... ... ...
95 0.443618 -1.473138 1.132161
96 -1.370215 -0.196425 -0.528401
97 1.062717 -0.997204 -1.666953
98 1.303512 0.699318 -0.863577
99 -0.109340 -1.330882 -1.455040
[100 rows x 3 columns]
Small simple function:
def ends(df, x=5):
return df.head(x).append(df.tail(x))
And use like so:
df = pd.DataFrame(np.random.rand(15,6))
ends(df,2)
I actually use this so much, I think it would be a great feature to add to pandas. (No features are to be added to pandas.DataFrame core API) I add it after import like so:
import pandas as pd
def ends(df, x=5):
return df.head(x).append(df.tail(x))
setattr(pd.DataFrame,'ends',ends)
Use like so:
import numpy as np
df = pd.DataFrame(np.random.rand(15,6))
df.ends(2)
You should use both head() and tail() for this purpose. I think the easiest way to do this is:
df.head(5).append(df.tail(5))
In Jupyter, expanding on #bolster's answer, we'll create a reusable convenience function:
def display_n(df,n):
with pd.option_context('display.max_rows',n*2):
display(df)
Then
display_n(df,2)
Returns
0 1 2
0 0.167961 -0.732745 0.952637
1 -0.050742 -0.421239 0.444715
... ... ... ...
98 0.085264 0.982093 -0.509356
99 -0.758963 -0.578267 -0.115865
(except as a nicely formatted HTML table)
when df is df = pd.DataFrame(np.random.randn(100,3))
Notes:
Of course you could make the same thing print as text by modifying display to print above.
On unix-like systems, you can the autoload the above function in all notebooks by placing it in a py or ipy file in ~/.ipython/profile_default/startup as described here.
If you want to keep it to just Pandas, you can use apply() to concatenate the head and tail:
import pandas as pd
from string import ascii_lowercase, ascii_uppercase
df = pd.DataFrame(
{"upper": list(ascii_uppercase), "lower": list(ascii_lowercase)}, index=range(1, 27)
)
df.apply(lambda x: pd.concat([x.head(2), x.tail(2)]))
upper lower
1 A a
2 B b
25 Y y
26 Z z
Associated with Linas Fx.
Defining below
pd.DataFrame.less = lambda df, n=10: df.head(n//2).append(df.tail(n//2))
then you can type only df.less()
It's same as type df.head().append(df.tail())
If you type df.less(2), the result is same as df.head(1).append(df.tail(1))
Combining #ic_fl2 and #watsonic to give the below in Jupyter:
def ends_attr():
def display_n(df,n):
with pd.option_context('display.max_rows',n*2):
display(df)
# set pd.DataFrame attribute where .ends runs display_n() function
setattr(pd.DataFrame,'ends',display_n)
ends_attr()
View first and last 3 rows of your df:
your_df.ends(3)
I like this because I can copy a single function and know I have everything I need to use the ends attribute.
I am trying to develop a new panda dataframe based on data I got from an existing dataframe and then taking into account the previously calculated value in the new dataframe.
As an example, here are two dataframes with the same size.
df1 = pd.DataFrame(np.random.randint(0,10, size = (5, 4)), columns=['1', '2', '3', '4'])
df2 = pd.DataFrame(np.zeros(df1.shape), index=df1.index, columns=df1.columns)
Then I created a list which starts as a starting basis for my second dataframe df2
L = [2,5,6,7]
df2.loc[0] = L
Then for the remaining rows of df2 I want to take the value from the previous time step (df2) and add the value of df1.
for i in df2.loc[1:]:
df2.ix[i] = df2.ix[i-1] + df1
As an example my dataframes should look like this:
>>> df1
1 2 3 4
0 4 6 0 6
1 7 0 7 9
2 9 1 9 9
3 5 2 3 6
4 0 3 2 9
>>> df2
1 2 3 4
0 2 5 6 7
1 9 5 13 16
2 18 6 22 25
3 23 8 25 31
4 23 11 27 40
I know there is something wrong with the indication of indexes in the for loop but I cannot figure out how the argument must be formulated. I would be very thankful for any help on this.
this is a simple cumsum.
df2 = df1.copy()
df2.loc[0] = [2,5,6,7]
desired_df = df2.cumsum()