Develop Reference

How to convert a nested list of keys to a dummies-like dataframe

How to convert following list to a pandas dataframe?
my_list = [["A","B","C"],["A","B","D"]]
And as an output I would like to have a dataframe like:
Index
A
B
C
D
1
1
1
1
0
2
1
1
0
1

You can craft Series and concatenate them:
my_list = [["A","B","C"],["A","B","D"]]
df = (pd.concat([pd.Series(1, index=l, name=i+1)
for i,l in enumerate(my_list)], axis=1)
.T
.fillna(0, downcast='infer') # optional
)
or with get_dummies:
df = pd.get_dummies(pd.DataFrame(my_list))
df = df.groupby(df.columns.str.split('_', 1).str[-1], axis=1).max()
output:
A B C D
1 1 1 1 0
2 1 1 0 1

I'm unsure how those two structures relate. The my_list is a list of two lists containing ["A","B","C"] and ["A", "B","D"].
If you want a data frame like the table you have, I would suggest making a dictionary of the values first, then converting it into a pandas dataframe.
my_dict = {"A":[1,1], "B":[1,1], "C": [1,0], "D":[0,1]}
my_df = pd.DataFrame(my_dict)
print(my_df)
Output:

Concat two Pandas DataFrame column with different length of index

How do I add a merge columns of Pandas dataframe to another dataframe while the new columns of data has less rows? Specifically I need to new column of data to be filled with NaN at the first few rows in the merged DataFrame instead of the last few rows. Please refer to the picture. Thanks.

Use:
df1 = pd.DataFrame({
'A':list('abcdef'),
'B':[4,5,4,5,5,4],
})
df2 = pd.DataFrame({
'SMA':list('rty')
})
df3 = df1.join(df2.set_index(df1.index[-len(df2):]))
Or:
df3 = pd.concat([df1, df2.set_index(df1.index[-len(df2):])], axis=1)
print (df3)
A B SMA
0 a 4 NaN
1 b 5 NaN
2 c 4 NaN
3 d 5 r
4 e 5 t
5 f 4 y
How it working:
First is selected index in df1 by length of df2 from back:
print (df1.index[-len(df2):])
RangeIndex(start=3, stop=6, step=1)
And then is overwrite existing values by DataFrame.set_index:
print (df2.set_index(df1.index[-len(df2):]))
SMA
3 r
4 t
5 y

Python: sample from dataframe, storing the non-sampled

I have a pandas DataFrame.
Say I want to sample two persons of each group, I use the following code to get a new dataframe:
sample_df = df.groupby("category").apply(lambda group_df: group_df.sample(2, random_state=1234)
I would like to create a dataframe where the non-sampled persons are stored.
The sample_df stil has the indices of the original df so I probably have to do something with that, but I'm not sure what...
Thanks in advance!

First add group_keys=False to groupby for avoid category to MultiIndex:
df = pd.DataFrame({
'A':list('abcdef'),
'B':[4,5,4,5,5,4],
'category':list('aaabbb')
})
sample_df = (df.groupby("category", group_keys=False)
.apply(lambda group_df: group_df.sample(2, random_state=1234)))
print(sample_df)
A B category
0 a 4 a
1 b 5 a
3 d 5 b
4 e 5 b
So possible filter original index values with boolean indexing by Index.isin and inverted mask by ~:
non_sample_df = df[~df.index.isin(sample_df.index)]
print(non_sample_df)
A B category
2 c 4 a
5 f 4 b

Python : How do you filter out columns from a dataset based on substring match in Column names

df_train = pd.read_csv('../xyz.csv')
headers = df_train.columns
I want to filter out those columns in headers which have _pct in their substring.

Use df.filter
df = pd.DataFrame({'a':[1,2,3], 'b_pct':[1,2,3],'c_pct':[1,2,3],'d':[1]*3})
print(df.filter(items=[i for i in df.columns if '_pct' not in i]))
## or as jezrael suggested
# print(df[[i for i in df.columns if '_pct' not in i]])
Output:
a d
0 1 1
1 2 1
2 3 1

Use:
#data from AkshayNevrekar answer
df = df.loc[:, ~df.columns.str.contains('_pct')]
print (df)
Filter solution is not trivial:
df = df.filter(regex=r'^(?!.*_pct).*$')
a d
0 1 1
1 2 1
2 3 1
Thank you, #IanS for another solutions:
df[df.columns.difference(df.filter(like='_pct').columns).tolist()]
df.drop(df.filter(like='_pct').columns, axis=1)

As df.columns returns a list of the column names, you can use list comprehension and build your new list with a simple condition:
new_headers = [x for x in headers if '_pct' not in x]

pandas convert grouped rows into columns

I have a dataframe such as:
label column1
a 1
a 2
b 6
b 4
I would like to make a dataframe with a new column, with the opposite value from column1 where the labels match. Such as:
label column1 column2
a 1 2
a 2 1
b 6 4
b 4 6
I know this is probably very simple to do with a groupby command but I've been searching and can't find anything.

The following uses groupby and apply and seems to work okay:
x = pd.DataFrame({ 'label': ['a','a','b','b'],
'column1': [1,2,6,4] })
y = x.groupby('label').apply(
lambda g: g.assign(column2 = np.asarray(g.column1[::-1])))
y = y.reset_index(drop=True) # optional: drop weird index
print(y)

you can try the code block below:
#create the Dataframe
df = pd.DataFrame({'label':['a','a','b','b'],
'column1':[1,2,6,4]})
#Group by label
a = df.groupby('label').first().reset_index()
b = df.groupby('label').last().reset_index()
#Concat those groups to create columns2
df2 = (pd.concat([b,a])
.sort_values(by='label')
.rename(columns={'column1':'column2'})
.reset_index()
.drop('index',axis=1))
#Merge with the original Dataframe
df = df.merge(df2,left_index=True,right_index=True,on='label')[['label','column1','column2']]
Hope this helps

Assuming their are only pairs of labels, you could use the following as well:
# Create dataframe
df = pd.DataFrame(data = {'label' :['a', 'a', 'b', 'b'],
'column1' :[1,2, 6,4]})
# iterate over dataframe, identify matching label and opposite value
for index, row in df.iterrows():
newvalue = int(df[(df.label == row.label) & (df.column1 != row.column1)].column1.values[0])
# set value to new column
df.set_value(index, 'column2', newvalue)
df.head()

You can use groupby with apply where create new Series with back order:
df['column2'] = df.groupby('label')["column1"] \
.apply(lambda x: pd.Series(x[::-1].values)).reset_index(drop=True)
print (df)
column1 label column2
0 1 a 2
1 2 a 1
2 6 b 4
3 4 b 6