pandas将分组行转换为列

Question

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.

Answer 1

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

Answer 2

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()

Answer 3

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)

Answer 4

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