Get rid of characters before a specific character and transform to float-CodePudding

I have a dataset like this:

    match                         one     ics      two
0   Juventus\n•\nSassuolo       1\n1.46 X\n5.25 2\n7.5
1   Sampdoria\n•\nAtalanta      1\n4.33 X\n4    2\n1.87
2   Udinese\n•\nHellas Verona   1\n2.52 X\n3.35 2\n3.1
3   Cagliari Calcio\n•\nRoma    1\n5.8  X\n4.33 2\n1.62
4   Empoli\n•\nInter            1\n7.2  X\n5.25 2\n1.47

I would like to get rid of the characters before and equal to n to convert the column one ics and two into float numbers. I do like this but it doesn't work

for index, row in matches.iterrows():
  matches.loc[index, 'one'] = matches.loc[index, 'one'].split('n', 1)[-1]
  matches.loc[index, 'ics'] = matches.loc[index, 'ics'].split('n', 1)[-1]
  matches.loc[index, 'two'] = matches.loc[index, 'two'].split('n', 1)[-1]
  matches['one'] = matches['one'].astype(float)
  matches['ics'] = matches['ics'].astype(float)
  matches['two'] = matches['two'].astype(float)

CodePudding user response：

If I understand your question, try:

temp = '2\n7.5'
    temp1 = temp.split('\n')
    print(temp1[0], ",", temp1[1], "or", temp1)

output:
2 , 7.5 or ['2', '7.5']

CodePudding user response：

Try with str.split and astype:

>>> df.set_index("match").apply(lambda x: (x.str.split().str[-1]).astype(float))

                            one   ics   two
match                                      
Juventus\n•\nSassuolo      1.46  5.25  7.50
Sampdoria\n•\nAtalanta     4.33  4.00  1.87
Udinese\n•\nHellas Verona  2.52  3.35  3.10
Cagliari Calcio\n•\nRoma   5.80  4.33  1.62
Empoli\n•\nInter           7.20  5.25  1.47