I have this DataFrame
lst = [['AAA',15,'BBB',20],['BBB',16,'AAA',12],['BBB',22,'CCC',15],['CCC',11,'AAA',31],['DDD',25,'EEE',35]]
df = pd.DataFrame(lst,columns = ['name1','val1','name2','val2'])
which looks like this
 name1   val1 name2 val2
0  AAA     15  BBB   20
1  BBB     16  AAA   12
2  BBB     22  CCC   15
3  CCC     11  AAA   31
4  DDD     25  EEE   35
I want this
 name1   val1 name2  val2
0  AAA     31  BBB    22
1  BBB     22  AAA    31
2  BBB     22  CCC    15
3  CCC     15  AAA    31
4  DDD     25  EEE    35
replaced all values with the maximum value. we choose the maximum value from both val1 and val2
if i do this i will get the maximum from only val1
df["val1"] = df.groupby("name1")["val1"].transform("max")
                To get the maximum value of each group, you can directly apply the pandas max() function to the selected column(s) from the result of pandas groupby.
How to groupby multiple columns in pandas DataFrame and compute multiple aggregations? groupby() can take the list of columns to group by multiple columns and use the aggregate functions to apply single or multiple aggregations at the same time.
Pandas replace multiple values in column replace. By using DataFrame. replace() method we will replace multiple values with multiple new strings or text for an individual DataFrame column. This method searches the entire Pandas DataFrame and replaces every specified value.
The Groupby Rolling function does not preserve the original index and so when dates are the same within the Group, it is impossible to know which index value it pertains to from the original dataframe.
Try using pd.wide_to_long to melt that dataframe into a long form, then use groupby with transform to find the max value.  Map that max value to 'name' and reshape back to four column (wide) dataframe:
df_long = pd.wide_to_long(df.reset_index(), ['name','val'], 'index', j='num',sep='',suffix='\d+')
mapper= df_long.groupby('name')['val'].max()
df_long['val'] = df_long['name'].map(mapper)
df_new = df_long.unstack()
df_new.columns = [f'{i}{j}' for i,j in df_new.columns]
df_new
Output:
      name1 name2  val1  val2
index                        
0       AAA   BBB    31    22
1       BBB   AAA    22    31
2       BBB   CCC    22    15
3       CCC   AAA    15    31
4       DDD   EEE    25    35
                        Borrow Scott's setting up
df_long = pd.wide_to_long(df.reset_index(), ['name','val'], 'index', j='num',sep='',suffix='\d+')
d = df_long.groupby('name')['val'].max()
df.loc[:,df.columns.str.startswith('val')]=df.loc[:,df.columns.str.startswith('name')].replace(d).values
df
Out[196]: 
  name1  val1 name2  val2
0   AAA    31   BBB    22
1   BBB    22   AAA    31
2   BBB    22   CCC    15
3   CCC    15   AAA    31
4   DDD    25   EEE    35
                        You can use lreshape (undocumented and ambiguous as to whether it's tested or will continue to remain) to get the long DataFrame, then map each pair of columns using the max.
names = df.columns[df.columns.str.startswith('name')]
vals = df.columns[df.columns.str.startswith('val')]
s = (pd.lreshape(df, groups={'name': names, 'val': vals})
       .groupby('name')['val'].max())
for n in names:
    df[n.replace('name', 'val')] = df[n].map(s)
  name1  val1 name2  val2
0   AAA    31   BBB    22
1   BBB    22   AAA    31
2   BBB    22   CCC    15
3   CCC    15   AAA    31
4   DDD    25   EEE    35
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