I'm currently working on some python dataframes over on pandas. And I'm not sure how this operation can be done. For example, I have an empty dataframe df and list of the following triples:
L = [(1,2,3), (2,5,4), (2,5,4), (3,2,0), (2,1,3)]
I wish to add all these triples into the dataframe df with columns ['id', 'a', 'b', 'c'] according to some constraint. The id is simply a counter that determines how many items have been added so far and a, b, and c are columns for the triples (but they would be commutative with each other). So the idea is to linearly traverse all items in L and then add each one to the df according to the restriction:
It is ok to add (1,2,3) since df is still empty. (id=0)
It is ok to add (2,5,4) since it or any of its permutation has not appeared yet in df. (id=1)
We then see (2,5,4) but this already exists in df, hence we cannot add it.
Next is (3,2,0) and we can clearly add this for the same reason as #2. (id=2)
Finally, it's (2,1,3). While this triple has not existed yet in df but since it's a permutation to an existing triplet in df (which is the (1,2,3)), then we cannot add it to df.
In the end, the final df should look something like this.
id a b c
0 1 2 3
1 2 5 4
2 3 2 0
Anyone knows how this can be done? My idea is to first curate an auxiliary list LL that would contain these "unique" triples and then just transform it into a pandas df. But I'm not sure if it's a fast and elegant efficient approach.
Fast solution
Create a numpy array from the list, then sort the array along axis=1 and use duplicated to create a boolean mask to identify dupes, then remove the duplicate rows from the array and create a new dataframe
a = np.array(L)
m = pd.DataFrame(np.sort(a, axis=1)).duplicated()
pd.DataFrame(a[~m], columns=['a', 'b', 'c'])
Result
a b c
0 1 2 3
1 2 5 4
2 3 2 0
You can use a dictionary comprehension with a frozenset of the tuple as key to eliminate the duplicated permutations, then feed the values to the DataFrame constructor:
L = [(1,2,3), (2,5,4), (2,5,4), (3,2,0), (2,1,3)]
df = pd.DataFrame({frozenset(t): t for t in L[::-1]}.values(),
columns=['a', 'b', 'c'])
output:
a b c
0 1 2 3
1 3 2 0
2 2 5 4
If order is important, you can use a set to collect the seen values instead:
seen = set()
df = pd.DataFrame([t for t in L if (f:=frozenset(t)) not in seen
and not seen.add(f)],
columns=['a', 'b', 'c'])
output:
a b c
0 1 2 3
1 2 5 4
2 3 2 0
handling duplicates values in the tuple
df = pd.DataFrame({tuple(sorted(t)): t
for t in L[::-1]}.values(),
columns=['a', 'b', 'c'])
If there are many columns, sorting becomes inefficient, then you can use a Counter:
from collections import Counter
df = pd.DataFrame({frozenset(Counter(t).items()): t
for t in L[::-1]}.values(),
columns=['a', 'b', 'c'])
pure pandas alternative:
You can do the same with pandas using loc and aggregation to set:
df = pd.DataFrame(L).loc[lambda d: ~d.agg(set, axis=1).duplicated()]
I want to merge several strings in a dataframe based on a groupedby in Pandas.
This is my code so far:
import pandas as pd
from io import StringIO
data = StringIO("""
"name1","hej","2014-11-01"
"name1","du","2014-11-02"
"name1","aj","2014-12-01"
"name1","oj","2014-12-02"
"name2","fin","2014-11-01"
"name2","katt","2014-11-02"
"name2","mycket","2014-12-01"
"name2","lite","2014-12-01"
""")
# load string as stream into dataframe
df = pd.read_csv(data,header=0, names=["name","text","date"],parse_dates=[2])
# add column with month
df["month"] = df["date"].apply(lambda x: x.month)
I want the end result to look like this:
I don't get how I can use groupby and apply some sort of concatenation of the strings in the column "text". Any help appreciated!
You can groupby the 'name' and 'month' columns, then call transform which will return data aligned to the original df and apply a lambda where we join the text entries:
In [119]:
df['text'] = df[['name','text','month']].groupby(['name','month'])['text'].transform(lambda x: ','.join(x))
df[['name','text','month']].drop_duplicates()
Out[119]:
name text month
0 name1 hej,du 11
2 name1 aj,oj 12
4 name2 fin,katt 11
6 name2 mycket,lite 12
I sub the original df by passing a list of the columns of interest df[['name','text','month']] here and then call drop_duplicates
EDIT actually I can just call apply and then reset_index:
In [124]:
df.groupby(['name','month'])['text'].apply(lambda x: ','.join(x)).reset_index()
Out[124]:
name month text
0 name1 11 hej,du
1 name1 12 aj,oj
2 name2 11 fin,katt
3 name2 12 mycket,lite
update
the lambda is unnecessary here:
In[38]:
df.groupby(['name','month'])['text'].apply(','.join).reset_index()
Out[38]:
name month text
0 name1 11 du
1 name1 12 aj,oj
2 name2 11 fin,katt
3 name2 12 mycket,lite
We can groupby the 'name' and 'month' columns, then call agg() functions of Panda’s DataFrame objects.
The aggregation functionality provided by the agg() function allows multiple statistics to be calculated per group in one calculation.
df.groupby(['name', 'month'], as_index = False).agg({'text': ' '.join})
The answer by EdChum provides you with a lot of flexibility but if you just want to concateate strings into a column of list objects you can also:
output_series = df.groupby(['name','month'])['text'].apply(list)
If you want to concatenate your "text" in a list:
df.groupby(['name', 'month'], as_index = False).agg({'text': list})
For me the above solutions were close but added some unwanted /n's and dtype:object, so here's a modified version:
df.groupby(['name', 'month'])['text'].apply(lambda text: ''.join(text.to_string(index=False))).str.replace('(\\n)', '').reset_index()
Please try this line of code : -
df.groupby(['name','month'])['text'].apply(','.join).reset_index()
Although, this is an old question. But just in case. I used the below code and it seems to work like a charm.
text = ''.join(df[df['date'].dt.month==8]['text'])
Thanks to all the other answers, the following is probably the most concise and feels more natural. Using df.groupby("X")["A"].agg() aggregates over one or many selected columns.
df = pandas.DataFrame({'A' : ['a', 'a', 'b', 'c', 'c'],
'B' : ['i', 'j', 'k', 'i', 'j'],
'X' : [1, 2, 2, 1, 3]})
A B X
a i 1
a j 2
b k 2
c i 1
c j 3
df.groupby("X", as_index=False)["A"].agg(' '.join)
X A
1 a c
2 a b
3 c
df.groupby("X", as_index=False)[["A", "B"]].agg(' '.join)
X A B
1 a c i i
2 a b j k
3 c j
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?
Edit 2017
As indicated in the comments and by #Alexander, currently the best method to add the values of a Series as a new column of a DataFrame could be using assign:
df1 = df1.assign(e=pd.Series(np.random.randn(sLength)).values)
Edit 2015
Some reported getting the SettingWithCopyWarning with this code.
However, the code still runs perfectly with the current pandas version 0.16.1.
>>> sLength = len(df1['a'])
>>> df1
a b c d
6 -0.269221 -0.026476 0.997517 1.294385
8 0.917438 0.847941 0.034235 -0.448948
>>> df1['e'] = pd.Series(np.random.randn(sLength), index=df1.index)
>>> df1
a b c d e
6 -0.269221 -0.026476 0.997517 1.294385 1.757167
8 0.917438 0.847941 0.034235 -0.448948 2.228131
>>> pd.version.short_version
'0.16.1'
The SettingWithCopyWarning aims to inform of a possibly invalid assignment on a copy of the Dataframe. It doesn't necessarily say you did it wrong (it can trigger false positives) but from 0.13.0 it let you know there are more adequate methods for the same purpose. Then, if you get the warning, just follow its advise: Try using .loc[row_index,col_indexer] = value instead
>>> df1.loc[:,'f'] = pd.Series(np.random.randn(sLength), index=df1.index)
>>> df1
a b c d e f
6 -0.269221 -0.026476 0.997517 1.294385 1.757167 -0.050927
8 0.917438 0.847941 0.034235 -0.448948 2.228131 0.006109
>>>
In fact, this is currently the more efficient method as described in pandas docs
Original answer:
Use the original df1 indexes to create the series:
df1['e'] = pd.Series(np.random.randn(sLength), index=df1.index)
This is the simple way of adding a new column: df['e'] = e
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.
Super simple column assignment
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
[]= is the way to go, but watch out!
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"
What actually is going on.
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
Side note
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.
Going around the problem
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".
Explicit way
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
Alternative way with 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
Note on 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.
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.
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())
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.
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.
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.
First create a python's list_of_e that has relevant data.
Use this:
df['e'] = list_of_e
To create an empty column
df['i'] = None
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.
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)
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
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)
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
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.
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 was looking for a general way of adding a column of numpy.nans to a dataframe without getting the dumb SettingWithCopyWarning.
From the following:
the answers here
this question about passing a variable as a keyword argument
this method for generating a numpy array of NaNs in-line
I came up with this:
col = 'column_name'
df = df.assign(**{col:numpy.full(len(df), numpy.nan)})
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
If you just need to create a new empty column then the shortest solution is:
df.loc[:, 'e'] = pd.Series()
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 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.
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
x=pd.DataFrame([1,2,3,4,5])
y=pd.DataFrame([5,4,3,2,1])
z=pd.concat([x,y],axis=1)
4 ways you can insert a new column to a pandas DataFrame
using simple assignment, insert(), assign() and Concat() methods.
import pandas as pd
df = pd.DataFrame({
'col_a':[True, False, False],
'col_b': [1, 2, 3],
})
print(df)
col_a col_b
0 True 1
1 False 2
2 False 3
Using simple assignment
ser = pd.Series(['a', 'b', 'c'], index=[0, 1, 2])
print(ser)
0 a
1 b
2 c
dtype: object
df['col_c'] = pd.Series(['a', 'b', 'c'], index=[1, 2, 3])
print(df)
col_a col_b col_c
0 True 1 NaN
1 False 2 a
2 False 3 b
Using assign()
e = pd.Series([1.0, 3.0, 2.0], index=[0, 2, 1])
ser = pd.Series(['a', 'b', 'c'], index=[0, 1, 2])
df.assign(colC=s.values, colB=e.values)
col_a col_b col_c
0 True 1.0 a
1 False 3.0 b
2 False 2.0 c
Using insert()
df.insert(len(df.columns), 'col_c', ser.values)
print(df)
col_a col_b col_c
0 True 1 a
1 False 2 b
2 False 3 c
Using concat()
ser = pd.Series(['a', 'b', 'c'], index=[10, 20, 30])
df = pd.concat([df, ser.rename('colC')], axis=1)
print(df)
col_a col_b col_c
0 True 1.0 NaN
1 False 2.0 NaN
2 False 3.0 NaN
10 NaN NaN a
20 NaN NaN b
30 NaN NaN c
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()
Whenever you add a Series object as new column to an existing DF, you need to make sure that they both have the same index.
Then add it to the DF
e_series = pd.Series([-0.335485, -1.166658,-0.385571])
print(e_series)
e_series.index = d_f.index
d_f['e'] = e_series
d_f
import pandas as pd
# Define a dictionary containing data
data = {'a': [0,0,0.671399,0.446172,0,0.614758],
'b': [0,0,0.101208,-0.243316,0,0.075793],
'c': [0,0,-0.181532,0.051767,0,-0.451460],
'd': [0,0,0.241273,1.577318,0,-0.012493]}
# Convert the dictionary into DataFrame
df = pd.DataFrame(data)
# Declare a list that is to be converted into a column
col_e = [-0.335485,-1.166658,-0.385571,0,0,0]
df['e'] = col_e
# add column 'e'
df['e'] = col_e
# Observe the result
df
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work