I have the following indexed DataFrame with named columns and rows not- continuous numbers:
a b c d
2 0.671399 0.101208 -0.181532 0.241273
3 0.446172 -0.243316 0.051767 1.577318
5 0.614758 0.075793 -0.451460 -0.012493
I would like to add a new column, 'e'
, to the existing data frame and do not want to change anything in the data frame (i.e., the new column always has the same length as the DataFrame).
0 -0.335485
1 -1.166658
2 -0.385571
dtype: float64
How can I add column e
to the above example?
This question is related to
python
pandas
dataframe
chained-assignment
Foolproof:
df.loc[:, 'NewCol'] = 'New_Val'
Example:
df = pd.DataFrame(data=np.random.randn(20, 4), columns=['A', 'B', 'C', 'D'])
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
3 -0.147354 0.778707 0.479145 2.284143
4 -0.529529 0.000571 0.913779 1.395894
5 2.592400 0.637253 1.441096 -0.631468
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
8 0.606985 -2.232903 -1.358107 -2.855494
9 -0.692013 0.671866 1.179466 -1.180351
10 -1.093707 -0.530600 0.182926 -1.296494
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
18 0.693458 0.144327 0.329500 -0.655045
19 0.104425 0.037412 0.450598 -0.923387
df.drop([3, 5, 8, 10, 18], inplace=True)
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
4 -0.529529 0.000571 0.913779 1.395894
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
9 -0.692013 0.671866 1.179466 -1.180351
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
19 0.104425 0.037412 0.450598 -0.923387
df.loc[:, 'NewCol'] = 0
df
A B C D NewCol
0 -0.761269 0.477348 1.170614 0.752714 0
1 1.217250 -0.930860 -0.769324 -0.408642 0
2 -0.619679 -1.227659 -0.259135 1.700294 0
4 -0.529529 0.000571 0.913779 1.395894 0
6 0.757178 0.240012 -0.553820 1.177202 0
7 -0.986128 -1.313843 0.788589 -0.707836 0
9 -0.692013 0.671866 1.179466 -1.180351 0
11 -0.143273 -0.503199 -1.328728 0.610552 0
12 -0.923110 -1.365890 -1.366202 -1.185999 0
13 -2.026832 0.273593 -0.440426 -0.627423 0
14 -0.054503 -0.788866 -0.228088 -0.404783 0
15 0.955298 -1.430019 1.434071 -0.088215 0
16 -0.227946 0.047462 0.373573 -0.111675 0
17 1.627912 0.043611 1.743403 -0.012714 0
19 0.104425 0.037412 0.450598 -0.923387 0
If you just need to create a new empty column then the shortest solution is:
df.loc[:, 'e'] = pd.Series()
One thing to note, though, is that if you do
df1['e'] = Series(np.random.randn(sLength), index=df1.index)
this will effectively be a left join on the df1.index. So if you want to have an outer join effect, my probably imperfect solution is to create a dataframe with index values covering the universe of your data, and then use the code above. For example,
data = pd.DataFrame(index=all_possible_values)
df1['e'] = Series(np.random.randn(sLength), index=df1.index)
Easiest ways:-
data['new_col'] = list_of_values
data.loc[ : , 'new_col'] = list_of_values
This way you avoid what is called chained indexing when setting new values in a pandas object. Click here to read further.
To create an empty column
df['i'] = None
If you want to set the whole new column to an initial base value (e.g. None
), you can do this: df1['e'] = None
This actually would assign "object" type to the cell. So later you're free to put complex data types, like list, into individual cells.
Let me just add that, just like for hum3, .loc
didn't solve the SettingWithCopyWarning
and I had to resort to df.insert()
. In my case false positive was generated by "fake" chain indexing dict['a']['e']
, where 'e'
is the new column, and dict['a']
is a DataFrame coming from dictionary.
Also note that if you know what you are doing, you can switch of the warning using
pd.options.mode.chained_assignment = None
and than use one of the other solutions given here.
Doing this directly via NumPy will be the most efficient:
df1['e'] = np.random.randn(sLength)
Note my original (very old) suggestion was to use map
(which is much slower):
df1['e'] = df1['a'].map(lambda x: np.random.random())
Before assigning a new column, if you have indexed data, you need to sort the index. At least in my case I had to:
data.set_index(['index_column'], inplace=True)
"if index is unsorted, assignment of a new column will fail"
data.sort_index(inplace = True)
data.loc['index_value1', 'column_y'] = np.random.randn(data.loc['index_value1', 'column_x'].shape[0])
To add a new column, 'e', to the existing data frame
df1.loc[:,'e'] = Series(np.random.randn(sLength))
I would like to add a new column, 'e', to the existing data frame and do not change anything in the data frame. (The series always got the same length as a dataframe.)
I assume that the index values in e
match those in df1
.
The easiest way to initiate a new column named e
, and assign it the values from your series e
:
df['e'] = e.values
assign (Pandas 0.16.0+)
As of Pandas 0.16.0, you can also use assign
, which assigns new columns to a DataFrame and returns a new object (a copy) with all the original columns in addition to the new ones.
df1 = df1.assign(e=e.values)
As per this example (which also includes the source code of the assign
function), you can also include more than one column:
df = pd.DataFrame({'a': [1, 2], 'b': [3, 4]})
>>> df.assign(mean_a=df.a.mean(), mean_b=df.b.mean())
a b mean_a mean_b
0 1 3 1.5 3.5
1 2 4 1.5 3.5
In context with your example:
np.random.seed(0)
df1 = pd.DataFrame(np.random.randn(10, 4), columns=['a', 'b', 'c', 'd'])
mask = df1.applymap(lambda x: x <-0.7)
df1 = df1[-mask.any(axis=1)]
sLength = len(df1['a'])
e = pd.Series(np.random.randn(sLength))
>>> df1
a b c d
0 1.764052 0.400157 0.978738 2.240893
2 -0.103219 0.410599 0.144044 1.454274
3 0.761038 0.121675 0.443863 0.333674
7 1.532779 1.469359 0.154947 0.378163
9 1.230291 1.202380 -0.387327 -0.302303
>>> e
0 -1.048553
1 -1.420018
2 -1.706270
3 1.950775
4 -0.509652
dtype: float64
df1 = df1.assign(e=e.values)
>>> df1
a b c d e
0 1.764052 0.400157 0.978738 2.240893 -1.048553
2 -0.103219 0.410599 0.144044 1.454274 -1.420018
3 0.761038 0.121675 0.443863 0.333674 -1.706270
7 1.532779 1.469359 0.154947 0.378163 1.950775
9 1.230291 1.202380 -0.387327 -0.302303 -0.509652
The description of this new feature when it was first introduced can be found here.
I was looking for a general way of adding a column of numpy.nan
s to a dataframe without getting the dumb SettingWithCopyWarning
.
From the following:
numpy
array of NaNs in-lineI came up with this:
col = 'column_name'
df = df.assign(**{col:numpy.full(len(df), numpy.nan)})
It seems that in recent Pandas versions the way to go is to use df.assign:
df1 = df1.assign(e=np.random.randn(sLength))
It doesn't produce SettingWithCopyWarning
.
If the column you are trying to add is a series variable then just :
df["new_columns_name"]=series_variable_name #this will do it for you
This works well even if you are replacing an existing column.just type the new_columns_name same as the column you want to replace.It will just overwrite the existing column data with the new series data.
I got the dreaded SettingWithCopyWarning
, and it wasn't fixed by using the iloc syntax. My DataFrame was created by read_sql from an ODBC source. Using a suggestion by lowtech above, the following worked for me:
df.insert(len(df.columns), 'e', pd.Series(np.random.randn(sLength), index=df.index))
This worked fine to insert the column at the end. I don't know if it is the most efficient, but I don't like warning messages. I think there is a better solution, but I can't find it, and I think it depends on some aspect of the index.
Note. That this only works once and will give an error message if trying to overwrite and existing column.
Note As above and from 0.16.0 assign is the best solution. See documentation http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.assign.html#pandas.DataFrame.assign
Works well for data flow type where you don't overwrite your intermediate values.
If we want to assign a scaler value eg: 10 to all rows of a new column in a df:
df = df.assign(new_col=lambda x:10) # x is each row passed in to the lambda func
df will now have new column 'new_col' with value=10 in all rows.
The following is what I did... But I'm pretty new to pandas and really Python in general, so no promises.
df = pd.DataFrame([[1, 2], [3, 4], [5,6]], columns=list('AB'))
newCol = [3,5,7]
newName = 'C'
values = np.insert(df.values,df.shape[1],newCol,axis=1)
header = df.columns.values.tolist()
header.append(newName)
df = pd.DataFrame(values,columns=header)
If you get the SettingWithCopyWarning
, an easy fix is to copy the DataFrame you are trying to add a column to.
df = df.copy()
df['col_name'] = values
A pandas dataframe is implemented as an ordered dict of columns.
This means that the __getitem__
[]
can not only be used to get a certain column, but __setitem__
[] =
can be used to assign a new column.
For example, this dataframe can have a column added to it by simply using the []
accessor
size name color
0 big rose red
1 small violet blue
2 small tulip red
3 small harebell blue
df['protected'] = ['no', 'no', 'no', 'yes']
size name color protected
0 big rose red no
1 small violet blue no
2 small tulip red no
3 small harebell blue yes
Note that this works even if the index of the dataframe is off.
df.index = [3,2,1,0]
df['protected'] = ['no', 'no', 'no', 'yes']
size name color protected
3 big rose red no
2 small violet blue no
1 small tulip red no
0 small harebell blue yes
However, if you have a pd.Series
and try to assign it to a dataframe where the indexes are off, you will run in to trouble. See example:
df['protected'] = pd.Series(['no', 'no', 'no', 'yes'])
size name color protected
3 big rose red yes
2 small violet blue no
1 small tulip red no
0 small harebell blue no
This is because a pd.Series
by default has an index enumerated from 0 to n. And the pandas [] =
method tries to be "smart"
When you use the [] =
method pandas is quietly performing an outer join or outer merge using the index of the left hand dataframe and the index of the right hand series. df['column'] = series
This quickly causes cognitive dissonance, since the []=
method is trying to do a lot of different things depending on the input, and the outcome cannot be predicted unless you just know how pandas works. I would therefore advice against the []=
in code bases, but when exploring data in a notebook, it is fine.
If you have a pd.Series
and want it assigned from top to bottom, or if you are coding productive code and you are not sure of the index order, it is worth it to safeguard for this kind of issue.
You could downcast the pd.Series
to a np.ndarray
or a list
, this will do the trick.
df['protected'] = pd.Series(['no', 'no', 'no', 'yes']).values
or
df['protected'] = list(pd.Series(['no', 'no', 'no', 'yes']))
But this is not very explicit.
Some coder may come along and say "Hey, this looks redundant, I'll just optimize this away".
Setting the index of the pd.Series
to be the index of the df
is explicit.
df['protected'] = pd.Series(['no', 'no', 'no', 'yes'], index=df.index)
Or more realistically, you probably have a pd.Series
already available.
protected_series = pd.Series(['no', 'no', 'no', 'yes'])
protected_series.index = df.index
3 no
2 no
1 no
0 yes
Can now be assigned
df['protected'] = protected_series
size name color protected
3 big rose red no
2 small violet blue no
1 small tulip red no
0 small harebell blue yes
df.reset_index()
Since the index dissonance is the problem, if you feel that the index of the dataframe should not dictate things, you can simply drop the index, this should be faster, but it is not very clean, since your function now probably does two things.
df.reset_index(drop=True)
protected_series.reset_index(drop=True)
df['protected'] = protected_series
size name color protected
0 big rose red no
1 small violet blue no
2 small tulip red no
3 small harebell blue yes
df.assign
While df.assign
make it more explicit what you are doing, it actually has all the same problems as the above []=
df.assign(protected=pd.Series(['no', 'no', 'no', 'yes']))
size name color protected
3 big rose red yes
2 small violet blue no
1 small tulip red no
0 small harebell blue no
Just watch out with df.assign
that your column is not called self
. It will cause errors. This makes df.assign
smelly, since there are these kind of artifacts in the function.
df.assign(self=pd.Series(['no', 'no', 'no', 'yes'])
TypeError: assign() got multiple values for keyword argument 'self'
You may say, "Well, I'll just not use self
then". But who knows how this function changes in the future to support new arguments. Maybe your column name will be an argument in a new update of pandas, causing problems with upgrading.
this is a special case of adding a new column to a pandas dataframe. Here, I am adding a new feature/column based on an existing column data of the dataframe.
so, let our dataFrame has columns 'feature_1', 'feature_2', 'probability_score' and we have to add a new_column 'predicted_class' based on data in column 'probability_score'.
I will use map() function from python and also define a function of my own which will implement the logic on how to give a particular class_label to every row in my dataFrame.
data = pd.read_csv('data.csv')
def myFunction(x):
//implement your logic here
if so and so:
return a
return b
variable_1 = data['probability_score']
predicted_class = variable_1.map(myFunction)
data['predicted_class'] = predicted_class
// check dataFrame, new column is included based on an existing column data for each row
data.head()
list_of_e
that has relevant data. df['e'] = list_of_e
If the data frame and Series object have the same index, pandas.concat
also works here:
import pandas as pd
df
# a b c d
#0 0.671399 0.101208 -0.181532 0.241273
#1 0.446172 -0.243316 0.051767 1.577318
#2 0.614758 0.075793 -0.451460 -0.012493
e = pd.Series([-0.335485, -1.166658, -0.385571])
e
#0 -0.335485
#1 -1.166658
#2 -0.385571
#dtype: float64
# here we need to give the series object a name which converts to the new column name
# in the result
df = pd.concat([df, e.rename("e")], axis=1)
df
# a b c d e
#0 0.671399 0.101208 -0.181532 0.241273 -0.335485
#1 0.446172 -0.243316 0.051767 1.577318 -1.166658
#2 0.614758 0.075793 -0.451460 -0.012493 -0.385571
In case they don't have the same index:
e.index = df.index
df = pd.concat([df, e.rename("e")], axis=1)
For the sake of completeness - yet another solution using DataFrame.eval() method:
Data:
In [44]: e
Out[44]:
0 1.225506
1 -1.033944
2 -0.498953
3 -0.373332
4 0.615030
5 -0.622436
dtype: float64
In [45]: df1
Out[45]:
a b c d
0 -0.634222 -0.103264 0.745069 0.801288
4 0.782387 -0.090279 0.757662 -0.602408
5 -0.117456 2.124496 1.057301 0.765466
7 0.767532 0.104304 -0.586850 1.051297
8 -0.103272 0.958334 1.163092 1.182315
9 -0.616254 0.296678 -0.112027 0.679112
Solution:
In [46]: df1.eval("e = @e.values", inplace=True)
In [47]: df1
Out[47]:
a b c d e
0 -0.634222 -0.103264 0.745069 0.801288 1.225506
4 0.782387 -0.090279 0.757662 -0.602408 -1.033944
5 -0.117456 2.124496 1.057301 0.765466 -0.498953
7 0.767532 0.104304 -0.586850 1.051297 -0.373332
8 -0.103272 0.958334 1.163092 1.182315 0.615030
9 -0.616254 0.296678 -0.112027 0.679112 -0.622436
to insert a new column at a given location (0 <= loc <= amount of columns) in a data frame, just use Dataframe.insert:
DataFrame.insert(loc, column, value)
Therefore, if you want to add the column e at the end of a data frame called df, you can use:
e = [-0.335485, -1.166658, -0.385571]
DataFrame.insert(loc=len(df.columns), column='e', value=e)
value can be a Series, an integer (in which case all cells get filled with this one value), or an array-like structure
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.insert.html
This is the simple way of adding a new column: df['e'] = e
Source: Stackoverflow.com