Mixing
Why can't you replace integers with lists using `replace` method - pandas
So let's say i have a pandas data-frame as below:
df=pd.DataFrame({'a':[1,2,3,0]})
So my goal is to replace 0 value with (empty list) in this dataframe, but i did:
print(df.replace(0,))
But it gives me an error:
TypeError: Invalid "to_replace" type: 'int'
I tried everything that's possible i.e:
df[df==0]=
etc...
But nothing works.
Desired output (in case for confusion):
a
0 1
1 2
2 3
3
python pandas dataframe replace types
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
add a comment |
So let's say i have a pandas data-frame as below:
df=pd.DataFrame({'a':[1,2,3,0]})
So my goal is to replace 0 value with (empty list) in this dataframe, but i did:
print(df.replace(0,))
But it gives me an error:
TypeError: Invalid "to_replace" type: 'int'
I tried everything that's possible i.e:
df[df==0]=
etc...
But nothing works.
Desired output (in case for confusion):
a
0 1
1 2
2 3
3
python pandas dataframe replace types
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
add a comment |
So let's say i have a pandas data-frame as below:
df=pd.DataFrame({'a':[1,2,3,0]})
So my goal is to replace 0 value with (empty list) in this dataframe, but i did:
print(df.replace(0,))
But it gives me an error:
TypeError: Invalid "to_replace" type: 'int'
I tried everything that's possible i.e:
df[df==0]=
etc...
But nothing works.
Desired output (in case for confusion):
a
0 1
1 2
2 3
3
python pandas dataframe replace types
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
So let's say i have a pandas data-frame as below:
df=pd.DataFrame({'a':[1,2,3,0]})
So my goal is to replace 0 value with (empty list) in this dataframe, but i did:
print(df.replace(0,))
But it gives me an error:
TypeError: Invalid "to_replace" type: 'int'
I tried everything that's possible i.e:
df[df==0]=
etc...
But nothing works.
Desired output (in case for confusion):
a
0 1
1 2
2 3
3
python pandas dataframe replace types
python pandas dataframe replace types
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
edited Dec 19 '18 at 0:16
U9-Forward
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
asked Nov 21 '18 at 10:34
U9-ForwardU9-Forward
15.2k41438
15.2k41438
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It is possible by list comprehension, but because mixed content - numeric with list it is not recommended:
df['a'] = [ if x == 0 else x for x in df.a]
print (df)
a
0 1
1 2
2 3
3
And replace all values in all columns:
df = df.applymap(lambda x: if x == 0 else x)
print (df)
a
0 1
1 2
2 3
3
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
1
Note that[] * len(df)may not work iflen(df) > 1. It will create a list containing the same instance of empty list rather thanlen(df)unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.
– Lie Ryan
Nov 21 '18 at 13:59
1
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
1
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
add a comment |
There are two issues here. First is pandas' quirkiness when dealing with lists. To replace values in a DataFrame with list you need to do something like this;
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
This creates n empty list based on the number of items that matches the criteria (df == 0)
The second issue is that your column is of integer type, and you can't store a list in an integer column. So before you can assign the list, you would first need to convert the column type to object first.
df = df.astype(object)
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
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
It is possible by list comprehension, but because mixed content - numeric with list it is not recommended:
df['a'] = [ if x == 0 else x for x in df.a]
print (df)
a
0 1
1 2
2 3
3
And replace all values in all columns:
df = df.applymap(lambda x: if x == 0 else x)
print (df)
a
0 1
1 2
2 3
3
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
1
Note that[] * len(df)may not work iflen(df) > 1. It will create a list containing the same instance of empty list rather thanlen(df)unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.
– Lie Ryan
Nov 21 '18 at 13:59
1
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
1
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
add a comment |
It is possible by list comprehension, but because mixed content - numeric with list it is not recommended:
df['a'] = [ if x == 0 else x for x in df.a]
print (df)
a
0 1
1 2
2 3
3
And replace all values in all columns:
df = df.applymap(lambda x: if x == 0 else x)
print (df)
a
0 1
1 2
2 3
3
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
1
Note that[] * len(df)may not work iflen(df) > 1. It will create a list containing the same instance of empty list rather thanlen(df)unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.
– Lie Ryan
Nov 21 '18 at 13:59
1
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
1
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
add a comment |
It is possible by list comprehension, but because mixed content - numeric with list it is not recommended:
df['a'] = [ if x == 0 else x for x in df.a]
print (df)
a
0 1
1 2
2 3
3
And replace all values in all columns:
df = df.applymap(lambda x: if x == 0 else x)
print (df)
a
0 1
1 2
2 3
3
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
It is possible by list comprehension, but because mixed content - numeric with list it is not recommended:
df['a'] = [ if x == 0 else x for x in df.a]
print (df)
a
0 1
1 2
2 3
3
And replace all values in all columns:
df = df.applymap(lambda x: if x == 0 else x)
print (df)
a
0 1
1 2
2 3
3
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
edited Nov 21 '18 at 14:00
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
answered Nov 21 '18 at 11:17
jezraeljezrael
334k25277352
334k25277352
1
Note that[] * len(df)may not work iflen(df) > 1. It will create a list containing the same instance of empty list rather thanlen(df)unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.
– Lie Ryan
Nov 21 '18 at 13:59
1
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
1
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
add a comment |
1
Note that[] * len(df)may not work iflen(df) > 1. It will create a list containing the same instance of empty list rather thanlen(df)unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.
– Lie Ryan
Nov 21 '18 at 13:59
1
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
1
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
1
1
Note that
[] * len(df) may not work if len(df) > 1. It will create a list containing the same instance of empty list rather than len(df) unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.– Lie Ryan
Nov 21 '18 at 13:59
Note that
[] * len(df) may not work if len(df) > 1. It will create a list containing the same instance of empty list rather than len(df) unique empty lists, which is probably not what you wanted. You'll need to use list comprehension to create new lists.– Lie Ryan
Nov 21 '18 at 13:59
1
1
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
Thanks so much, i appreciate it, came back and accepted this!!!
– U9-Forward
Nov 22 '18 at 8:10
1
1
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
@U9-Forward - Thank you very much!
– jezrael
Nov 22 '18 at 8:10
add a comment |
There are two issues here. First is pandas' quirkiness when dealing with lists. To replace values in a DataFrame with list you need to do something like this;
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
This creates n empty list based on the number of items that matches the criteria (df == 0)
The second issue is that your column is of integer type, and you can't store a list in an integer column. So before you can assign the list, you would first need to convert the column type to object first.
df = df.astype(object)
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
add a comment |
There are two issues here. First is pandas' quirkiness when dealing with lists. To replace values in a DataFrame with list you need to do something like this;
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
This creates n empty list based on the number of items that matches the criteria (df == 0)
The second issue is that your column is of integer type, and you can't store a list in an integer column. So before you can assign the list, you would first need to convert the column type to object first.
df = df.astype(object)
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
add a comment |
There are two issues here. First is pandas' quirkiness when dealing with lists. To replace values in a DataFrame with list you need to do something like this;
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
This creates n empty list based on the number of items that matches the criteria (df == 0)
The second issue is that your column is of integer type, and you can't store a list in an integer column. So before you can assign the list, you would first need to convert the column type to object first.
df = df.astype(object)
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
There are two issues here. First is pandas' quirkiness when dealing with lists. To replace values in a DataFrame with list you need to do something like this;
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
This creates n empty list based on the number of items that matches the criteria (df == 0)
The second issue is that your column is of integer type, and you can't store a list in an integer column. So before you can assign the list, you would first need to convert the column type to object first.
df = df.astype(object)
df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
answered Nov 21 '18 at 13:50
Lie RyanLie Ryan
45k968122
45k968122
add a comment |
add a comment |
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