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Pandas: Get top n columns based on a row values

Having a dataframe with a single row, I need to filter it into a smaller one with filtered columns based on a value in a row.
What's the most effective way?
df = pd.DataFrame({'a':[1], 'b':[10], 'c':[3], 'd':[5]})
a
b
c
d
1
10
3
5
For example top-3 features:
b
c
d
10
3
5

Use sorting per row and select first 3 values:
df1 = df.sort_values(0, axis=1, ascending=False).iloc[:, :3]
print (df1)
b d c
0 10 5 3
Solution with Series.nlargest:
df1 = df.iloc[0].nlargest(3).to_frame().T
print (df1)
b d c
0 10 5 3

You can transpose T, and use nlargest():
new = df.T.nlargest(columns = 0, n = 3).T
print(new)
b d c
0 10 5 3

You can use np.argsort to get the solution. This Numpy method, in the below code, gives the indices of the column values in descending order. Then slicing selects the largest n values' indices.
import pandas as pd
import numpy as np
# Your dataframe
df = pd.DataFrame({'a':[1], 'b':[10], 'c':[3], 'd':[5]})
# Pick the number n to find n largest values
nlargest = 3
# Get the order of the largest value columns by their indices
order = np.argsort(-df.values, axis=1)[:, :nlargest]
# Find the columns with the largest values
top_features = df.columns[order].tolist()[0]
# Filter the dateframe by the columns
top_features_df = df[top_features]
top_features_df
output:
b d c
0 10 5 3

Python Pandas pandas correlation one column vs all

I'm trying to get the correlation between a single column and the rest of the numerical columns of the dataframe, but I'm stuck.
I'm trying with this:
corr = IM['imdb_score'].corr(IM)
But I get the error
operands could not be broadcast together with shapes
which I assume is because I'm trying to find a correlation between a vector (my imdb_score column) with the dataframe of several columns.
How can this be fixed?

The most efficient method it to use corrwith.
Example:
df.corrwith(df['A'])
Setup of example data:
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(10, size=(5, 5)), columns=list('ABCDE'))
# A B C D E
# 0 7 2 0 0 0
# 1 4 4 1 7 2
# 2 6 2 0 6 6
# 3 9 8 0 2 1
# 4 6 0 9 7 7
output:
A 1.000000
B 0.526317
C -0.209734
D -0.720400
E -0.326986
dtype: float64

I think you can you just use .corr which returns all correlations between all columns and then select just the column you are interested in.
So, something like
IM.corr()['imbd_score']
should work.

Rather than calculating all correlations and keeping the ones of interest, it can be computationally more efficient to compute the subset of interesting correlations:
import pandas as pd
df = pd.DataFrame()
df['a'] = range(10)
df['b'] = range(10)
df['c'] = range(10)
pd.DataFrame([[c, df['a'].corr(df[c])] for c in df.columns if c!='a'], columns=['var', 'corr'])

How to transform the result of a Pandas `GROUPBY` function to the original dataframe

Suppose I have a Pandas DataFrame with 6 columns and a custom function that takes counts of the elements in 2 or 3 columns and produces a boolean output. When a groupby object is created from the original dataframe and the custom function is applied df.groupby('col1').apply(myfunc), the result is a series whose length is equal to the number of categories of col1. How do I expand this output to match the length of the original dataframe? I tried transform, but was not able to use the custom function myfunc with it.
EDIT:
Here is an example code:
A = pd.DataFrame({'X':['a','b','c','a','c'], 'Y':['at','bt','ct','at','ct'], 'Z':['q','q','r','r','s']})
print (A)
def myfunc(df):
return ((df['Z'].nunique()>=2) and (df['Y'].nunique()<2))
A.groupby('X').apply(myfunc)
I would like to expand this output as a new column Result such that where there is a in column X, the Result will be True.

You can map the groupby back to the original dataframe
A['Result'] = A['X'].map(A.groupby('X').apply(myfunc))
Result would look like:
X Y Z Result
0 a at q True
1 b bt q False
2 c ct r True
3 a at r True
4 c ct s True

My solution may not be the best one, which uses a loop, but it's pretty good I think.
The core idea is you can traverse all the sub-dataframe (gdf) by for i, gdf in gp. Then add the column result (in my example it is c) for each sub-dataframe. Finally concat all the sub-dataframe into one.
Here is an example:
import pandas as pd
df = pd.DataFrame({'a':[1,2,1,2],'b':['a','b','c','d']})
gp = df.groupby('a') # group
s = gp.apply(sum)['a'] # apply a func
adf = []
# then create a new dataframe
for i, gdf in gp:
tdf = gdf.copy()
tdf.loc[:,'c'] = s.loc[i]
adf.append(tdf)
pd.concat(adf)
from:
a b
0 1 a
1 2 b
2 1 c
3 2 d
to:
a b c
0 1 a 2
2 1 c 2
1 2 b 4
3 2 d 4

Unpack DataFrame with tuple entries into separate DataFrames

I wrote a small class to compute some statistics through bootstrap without replacement. For those not familiar with this technique, you get n random subsamples of some data, compute the desired statistic (lets say the median) on each subsample, and then compare the values across subsamples. This allows you to get a measure of variance on the obtained median over the dataset.
I implemented this in a class but reduced it to a MWE given by the following function
import numpy as np
import pandas as pd
def bootstrap_median(df, n=5000, fraction=0.1):
if isinstance(df, pd.DataFrame):
columns = df.columns
else:
columns = None
# Get the values as a ndarray
arr = np.array(df.values)
# Get the bootstrap sample through random permutations
sample_len = int(len(arr)*fraction)
if sample_len<1:
sample_len = 1
sample = []
for n_sample in range(n):
sample.append(arr[np.random.permutation(len(arr))[:sample_len]])
sample = np.array(sample)
# Compute the median on each sample
temp = np.median(sample, axis=1)
# Get the mean and std of the estimate across samples
m = np.mean(temp, axis=0)
s = np.std(temp, axis=0)/np.sqrt(len(sample))
# Convert output to DataFrames if necesary and return
if columns:
m = pd.DataFrame(data=m[None, ...], columns=columns)
s = pd.DataFrame(data=s[None, ...], columns=columns)
return m, s
This function returns the mean and standard deviation across the medians computed on each bootstrap sample.
Now consider this example DataFrame
data = np.arange(20)
group = np.tile(np.array([1, 2]).reshape(-1,1), (1,10)).flatten()
df = pd.DataFrame.from_dict({'data': data, 'group': group})
print(df)
print(bootstrap_median(df['data']))
this prints
data group
0 0 1
1 1 1
2 2 1
3 3 1
4 4 1
5 5 1
6 6 1
7 7 1
8 8 1
9 9 1
10 10 2
11 11 2
12 12 2
13 13 2
14 14 2
15 15 2
16 16 2
17 17 2
18 18 2
19 19 2
(9.5161999999999995, 0.056585753613431718)
So far so good because bootstrap_median returns a tuple of two elements. However, if I do this after a groupby
In: df.groupby('group')['data'].apply(bootstrap_median)
Out:
group
1 (4.5356, 0.0409710449952)
2 (14.5006, 0.0403772204095)
The values inside each cell are tuples, as one would expect from apply. I can unpack the result into two DataFrame's by iterating over elements like this:
index = []
data1 = []
data2 = []
for g, (m, s) in out.iteritems():
index.append(g)
data1.append(m)
data2.append(s)
dfm = pd.DataFrame(data=data1, index=index, columns=['E[median]'])
dfm.index.name = 'group'
dfs = pd.DataFrame(data=data2, index=index, columns=['std[median]'])
dfs.index.name = 'group'
thus
In: dfm
Out:
E[median]
group
1 4.5356
2 14.5006
In: dfs
Out:
std[median]
group
1 0.0409710449952
2 0.0403772204095
This is a bit cumbersome and my question is if there is a more pandas native way to "unpack" a dataframe whose values are tuples into separate DataFrame's
This question seemed related but it concerned string regex replacements and not unpacking true tuples.

I think you need change:
return m, s
to:
return pd.Series([m, s], index=['m','s'])
And then get:
df1 = df.groupby('group')['data'].apply(bootstrap_median)
print (df1)
group
1 m 4.480400
s 0.040542
2 m 14.565200
s 0.040373
Name: data, dtype: float64
So is possible select by xs:
print (df1.xs('s', level=1))
group
1 0.040542
2 0.040373
Name: data, dtype: float64
print (df1.xs('m', level=1))
group
1 4.4804
2 14.5652
Name: data, dtype: float64
Also if need one column DataFrame add to_frame:
df1 = df.groupby('group')['data'].apply(bootstrap_median).to_frame()
print (df1)
data
group
1 m 4.476800
s 0.041100
2 m 14.468400
s 0.040719
print (df1.xs('s', level=1))
data
group
1 0.041100
2 0.040719
print (df1.xs('m', level=1))
data
group
1 4.4768
2 14.4684

add columns different length pandas

I have a problem with adding columns in pandas.
I have DataFrame, dimensional is nxk. And in process I wiil need add columns with dimensional mx1, where m = [1,n], but I don't know m.
When I try do it:
df['Name column'] = data
# type(data) = list
result:
AssertionError: Length of values does not match length of index
Can I add columns with different length?

If you use accepted answer, you'll lose your column names, as shown in the accepted answer example, and described in the documentation (emphasis added):
The resulting axis will be labeled 0, ..., n - 1. This is useful if you are concatenating objects where the concatenation axis does not have meaningful indexing information.
It looks like column names ('Name column') are meaningful to the Original Poster / Original Question.
To save column names, use pandas.concat, but don't ignore_index (default value of ignore_index is false; so you can omit that argument altogether). Continue to use axis=1:
import pandas
# Note these columns have 3 rows of values:
original = pandas.DataFrame({
'Age':[10, 12, 13],
'Gender':['M','F','F']
})
# Note this column has 4 rows of values:
additional = pandas.DataFrame({
'Name': ['Nate A', 'Jessie A', 'Daniel H', 'John D']
})
new = pandas.concat([original, additional], axis=1)
# Identical:
# new = pandas.concat([original, additional], ignore_index=False, axis=1)
print(new.head())
# Age Gender Name
#0 10 M Nate A
#1 12 F Jessie A
#2 13 F Daniel H
#3 NaN NaN John D
Notice how John D does not have an Age or a Gender.

Use concat and pass axis=1 and ignore_index=True:
In [38]:
import numpy as np
df = pd.DataFrame({'a':np.arange(5)})
df1 = pd.DataFrame({'b':np.arange(4)})
print(df1)
df
b
0 0
1 1
2 2
3 3
Out[38]:
a
0 0
1 1
2 2
3 3
4 4
In [39]:
pd.concat([df,df1], ignore_index=True, axis=1)
Out[39]:
0 1
0 0 0
1 1 1
2 2 2
3 3 3
4 4 NaN

We can add the different size of list values to DataFrame.
Example
a = [0,1,2,3]
b = [0,1,2,3,4,5,6,7,8,9]
c = [0,1]
Find the Length of all list
la,lb,lc = len(a),len(b),len(c)
# now find the max
max_len = max(la,lb,lc)
Resize all according to the determined max length (not in this example
if not max_len == la:
a.extend(['']*(max_len-la))
if not max_len == lb:
b.extend(['']*(max_len-lb))
if not max_len == lc:
c.extend(['']*(max_len-lc))
Now the all list is same length and create dataframe
pd.DataFrame({'A':a,'B':b,'C':c})
Final Output is
A B C
0 1 0 1
1 2 1
2 3 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9

I had the same issue, two different dataframes and without a common column. I just needed to put them beside each other in a csv file.
Merge:
In this case, "merge" does not work; even adding a temporary column to both dfs and then dropping it. Because this method makes both dfs with the same length. Hence, it repeats the rows of the shorter dataframe to match the longer dataframe's length.
Concat:
The idea of The Red Pea didn't work for me. It just appended the shorter df to the longer one (row-wise) while leaving an empty column (NaNs) above the shorter df's column.
Solution: You need to do the following:
df1 = df1.reset_index()
df2 = df2.reset_index()
df = [df1, df2]
df_final = pd.concat(df, axis=1)
df_final.to_csv(filename, index=False)
This way, you'll see your dfs besides each other (column-wise), each of which with its own length.

If somebody like to replace a specific column of a different size instead of adding it.
Based on this answer, I use a dict as an intermediate type.
Create Pandas Dataframe with different sized columns
If the column to be inserted is not a list but already a dict, the respective line can be omitted.
def fill_column(dataframe: pd.DataFrame, list: list, column: str):
dict_from_list = dict(enumerate(list)) # create enumertable object from list and create dict
dataFrame_asDict = dataframe.to_dict() # Get DataFrame as Dict
dataFrame_asDict[column] = dict_from_list # Assign specific column
return pd.DataFrame.from_dict(dataFrame_asDict, orient='index').T # Create new DataSheet from Dict and return it