pandas dataframe apply lambda index error
I have the following code
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1])
where df2 is:
TaxAccNo2
0 00001379.1
1 00182218
When I run the code I get
TaxAccNo2 TaxAccNo4
0 00001379.1 00001379
1 00182218 00182218
and IndexError: list index out of range for TaxAccNo3,
TaxAccNo2 TaxAccNo4 TaxAccNo3
0 00001379.1 00001379 1
1 00182218 00182218
How do I fix my code to produce that output? I'm assuming its giving me the error because Index 1 doesn't have '.' but I'm not sure how to fix that.
python pandas dataframe
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
asked Nov 21 '18 at 1:09
AndyAndy
437
add a comment |
I have the following code
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1])
where df2 is:
TaxAccNo2
0 00001379.1
1 00182218
When I run the code I get
TaxAccNo2 TaxAccNo4
0 00001379.1 00001379
1 00182218 00182218
and IndexError: list index out of range for TaxAccNo3,
TaxAccNo2 TaxAccNo4 TaxAccNo3
0 00001379.1 00001379 1
1 00182218 00182218
How do I fix my code to produce that output? I'm assuming its giving me the error because Index 1 doesn't have '.' but I'm not sure how to fix that.
python pandas dataframe
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
asked Nov 21 '18 at 1:09
AndyAndy
437
add a comment |
I have the following code
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1])
where df2 is:
TaxAccNo2
0 00001379.1
1 00182218
When I run the code I get
TaxAccNo2 TaxAccNo4
0 00001379.1 00001379
1 00182218 00182218
and IndexError: list index out of range for TaxAccNo3,
TaxAccNo2 TaxAccNo4 TaxAccNo3
0 00001379.1 00001379 1
1 00182218 00182218
How do I fix my code to produce that output? I'm assuming its giving me the error because Index 1 doesn't have '.' but I'm not sure how to fix that.
python pandas dataframe
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
asked Nov 21 '18 at 1:09
AndyAndy
437
I have the following code
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1])
where df2 is:
TaxAccNo2
0 00001379.1
1 00182218
When I run the code I get
TaxAccNo2 TaxAccNo4
0 00001379.1 00001379
1 00182218 00182218
and IndexError: list index out of range for TaxAccNo3,
TaxAccNo2 TaxAccNo4 TaxAccNo3
0 00001379.1 00001379 1
1 00182218 00182218
How do I fix my code to produce that output? I'm assuming its giving me the error because Index 1 doesn't have '.' but I'm not sure how to fix that.
python pandas dataframe
python pandas dataframe
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
asked Nov 21 '18 at 1:09
AndyAndy
437
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
asked Nov 21 '18 at 1:09
AndyAndy
437
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
edited Nov 21 '18 at 1:11
eyllanesc
76k103156
76k103156
asked Nov 21 '18 at 1:09
AndyAndy
437
asked Nov 21 '18 at 1:09
AndyAndy
437
asked Nov 21 '18 at 1:09
AndyAndy
437
437
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
As you said, the problem is that "00182218".split(".") doesn't have a [1] index, since it's the list ["00182218"].
A simple solution without affecting too much the code is to use a ... if ... else ... ternary operator:
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if '.' in x else '')
Where the last '' is an empty string, the value with which you will fill 'TaxAccNo3' if 'TaxAccNo2' doesn't have a dot (you can replace it if you want other behaviour).
The semantic is: put x.split('.')[1] in df2['TaxAccNo3'] if x contains a dot, otherwise put an empty string.
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
add a comment |
Hy, I was reviewing your code, the problem is that when you use the method split() in a string the returned object is a list, and this is causing the index error, as you pointed. The solution I encountered if very simple, use a conditional in your code to prevent it from calling this index for shorter lists as follows. Hope it helps.
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if len(x.split('.'))>1 else x)
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
As you said, the problem is that "00182218".split(".") doesn't have a [1] index, since it's the list ["00182218"].
A simple solution without affecting too much the code is to use a ... if ... else ... ternary operator:
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if '.' in x else '')
Where the last '' is an empty string, the value with which you will fill 'TaxAccNo3' if 'TaxAccNo2' doesn't have a dot (you can replace it if you want other behaviour).
The semantic is: put x.split('.')[1] in df2['TaxAccNo3'] if x contains a dot, otherwise put an empty string.
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
add a comment |
As you said, the problem is that "00182218".split(".") doesn't have a [1] index, since it's the list ["00182218"].
A simple solution without affecting too much the code is to use a ... if ... else ... ternary operator:
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if '.' in x else '')
Where the last '' is an empty string, the value with which you will fill 'TaxAccNo3' if 'TaxAccNo2' doesn't have a dot (you can replace it if you want other behaviour).
The semantic is: put x.split('.')[1] in df2['TaxAccNo3'] if x contains a dot, otherwise put an empty string.
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
add a comment |
As you said, the problem is that "00182218".split(".") doesn't have a [1] index, since it's the list ["00182218"].
A simple solution without affecting too much the code is to use a ... if ... else ... ternary operator:
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if '.' in x else '')
Where the last '' is an empty string, the value with which you will fill 'TaxAccNo3' if 'TaxAccNo2' doesn't have a dot (you can replace it if you want other behaviour).
The semantic is: put x.split('.')[1] in df2['TaxAccNo3'] if x contains a dot, otherwise put an empty string.
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
As you said, the problem is that "00182218".split(".") doesn't have a [1] index, since it's the list ["00182218"].
A simple solution without affecting too much the code is to use a ... if ... else ... ternary operator:
df2['TaxAccNo4'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[0])
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if '.' in x else '')
Where the last '' is an empty string, the value with which you will fill 'TaxAccNo3' if 'TaxAccNo2' doesn't have a dot (you can replace it if you want other behaviour).
The semantic is: put x.split('.')[1] in df2['TaxAccNo3'] if x contains a dot, otherwise put an empty string.
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
answered Nov 21 '18 at 1:19
Julian PellerJulian Peller
8941511
8941511
add a comment |
add a comment |
Hy, I was reviewing your code, the problem is that when you use the method split() in a string the returned object is a list, and this is causing the index error, as you pointed. The solution I encountered if very simple, use a conditional in your code to prevent it from calling this index for shorter lists as follows. Hope it helps.
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if len(x.split('.'))>1 else x)
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
add a comment |
Hy, I was reviewing your code, the problem is that when you use the method split() in a string the returned object is a list, and this is causing the index error, as you pointed. The solution I encountered if very simple, use a conditional in your code to prevent it from calling this index for shorter lists as follows. Hope it helps.
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if len(x.split('.'))>1 else x)
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
add a comment |
Hy, I was reviewing your code, the problem is that when you use the method split() in a string the returned object is a list, and this is causing the index error, as you pointed. The solution I encountered if very simple, use a conditional in your code to prevent it from calling this index for shorter lists as follows. Hope it helps.
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if len(x.split('.'))>1 else x)
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
Hy, I was reviewing your code, the problem is that when you use the method split() in a string the returned object is a list, and this is causing the index error, as you pointed. The solution I encountered if very simple, use a conditional in your code to prevent it from calling this index for shorter lists as follows. Hope it helps.
df2['TaxAccNo3'] = df2['TaxAccNo2'].apply(lambda x: x.split('.')[1] if len(x.split('.'))>1 else x)
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
answered Nov 21 '18 at 1:51
Pedro MorescoPedro Moresco
214
214
add a comment |
add a comment |
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