This is what my main dataframe looks like:
Group IDs New ID
1 [N23,N1,N12] N102
2 [N134,N100] N501
I have another dataframe that has all the required ID info in an unordered manner:
ID Name Age
N1 Milo 5
N23 Mark 21
N11 Jacob 22
I would like to modify the original dataframe such that all IDs are replaced with their respective names obtained from the other file. So that the dataframe has only names and no IDs and looks like this:
Group IDs New ID
1 [Mark,Silo,Bond] Niki
2 [Troy,Fangio] Kvyat
Thanks in advance
IIUC you can .explode your lists, replace values with .map and regroup them with .groupby
df['ID'] = (df.ID.explode()
.map(df1.set_index('ID')['Name'])
.groupby(level=0).agg(list)
)
If New ID column is not a list, you can use only .map()
df['New ID'] = df['New ID'].map(df1.set_index('ID')['Name'])
you can try making a dict from your second DF and then replacing on the first using regex patterns (no need to fully understand it, check the comments bellow):
ps: since you didn't provide the full df with the codes, I created with some of them, that's why the print() won't replace all the results.
import pandas as pd
# creating dummy dfs
df1 = pd.DataFrame({"Group":[1,2], "IDs":["[N23,N1,N12]", "[N134,N100]"], "New ID":["N102", "N501"] })
df2 = pd.DataFrame({"ID":['N1', "N23", "N11", "N100"], "Name":["Milo", "Mark", "Jacob", "Silo"], "Age":[5,21,22, 44]})
# Create the unique dict we're using regex patterns to make exact match
dict_replace = df2.set_index("ID")['Name'].to_dict()
# 'f' before string means fstrings and 'r' means to interpret it as regex
# the \b is a regex pattern that it sinalizes the begining and end of the match
## so that if you're searching for N1, it won't match if it is N11
dict_replace = {fr"\b{k}\b":v for k, v in dict_replace.items()}
# Replacing on original where you want it
df1['IDs'].replace(dict_replace, regex=True, inplace=True)
print(df1['IDs'].tolist())
# >>> ['[Mark,Milo,N12]', '[N134,Silo]']
Please note the change in my dataframes. In your sample data, the IDs in df that do not exists in df1 IDs. I altered my df to ensure only IDs in df1 were represented. I use the following df
print(df)
Group IDs New
0 1 [N23,N1,N11] N102
1 2 [N11,N23] N501
print(df1)
ID Name Age
0 N1 Milo 5
1 N23 Mark 21
2 N11 Jacob 22
Solution
dict df1.Id and df.Name and map to an exploded df.IDs. Add the result to list.
df['IDs'] = df['IDs'].str.strip('[]')#Strip corner brackets
df['IDs'] = df['IDs'].str.split(',')#Reconstruct list, this was done because for some reason I couldnt explode list
#df.explode list and map df1 to df and add to list
df.explode('IDs').groupby('Group')['IDs'].apply(lambda x:(x.map(dict(zip(df1.ID,df1.Name)))).tolist()).reset_index()
Group IDs
0 1 [Mark, Milo, Jacob]
1 2 [Jacob, Mark]
What I want to achieve.
I have two data frames. DF1 and DF2. Both are being read from different excel file.
DF1 has 9 columns and 3000 rows, of which one of the column name is "Code Group".
DF2 has 2 columns and 20 rows, of which one of the column name is also "Code Group". In this same dataframe another column "Code Management Method" gives the explanation of code group. For eg. H001 is Treated at recyclable, H002 is Treated as landfill.
What happens
When I use the command data = pd.merge(DF1,DF2, on='Code Group') I only get 10 column names but no rows underneath.
What I expect
I would want DF1 and DF2 to be merged and wherever Code Group number is same Code Management Method to be pasted for explanation.
Additional information
Following are datatype for DF1
Entity object
Address object
State object
Site object
Disposal Facility object
Pounds float64
Waste Description object
Shipment Date datetime64[ns]
Code Group object
FollOwing are datatype for DF2
Code Management Method object
Code Group object
What I tried
I tried to follow the suggestions from similar post on SO that the datatypes on both sides should be same and Code Group here both are objects so don't know what's the issue. I also tried Concat function.
Code
import pandas as pd
from pandas import ExcelWriter
from pandas import ExcelFile
CH = "C:\Python\Waste\Shipment.xls"
Code = "C:\Python\Waste\Code.xlsx"
Data = pd.read_excel(Code)
data1 = pd.read_excel(CH)
data1.rename(columns={'generator_name':'Entity','generator_address':'Address', 'Site_City':'Site','final_disposal_facility_name':'Disposal Facility', 'wst_dscrpn':'Waste Description', 'drum_wgt':'Pounds', 'wst_dscrpn' : 'Waste Description', 'genrtr_sgntr_dt':'Shipment Date','generator_state': 'State','expected_disposal_management_methodcode':'Code Group'},
inplace=True)
data2 = data1[['Entity','Address','State','Site','Disposal Facility','Pounds','Waste Description','Shipment Date','Code Group']]
data2
merged = data2.merge(Data, on='Code Group')
Getting a Warning
C:\Anaconda\lib\site-packages\pandas\core\generic.py:5890: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame
See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
self._update_inplace(new_data)
import pandas as pd
df1 = pd.DataFrame({'Region': [1,2,3],
'zipcode':[12345,23456,34567]})
df2 = pd.DataFrame({'ZipCodeLowerBound': [10000,20000,30000],
'ZipCodeUpperBound': [19999,29999,39999],
'Region': [1,2,3]})
df1.merge(df2, on='Region')
this is how the example is given, and the result for this is:
Region zipcode
0 1 12345
1 2 23456
2 3 34567
Region ZipCodeLowerBound ZipCodeUpperBound
0 1 10000 19999
1 2 20000 29999
2 3 30000 39999
and that thing will result in
Region zipcode ZipCodeLowerBound ZipCodeUpperBound
0 1 12345 10000 19999
1 2 23456 20000 29999
2 3 34567 30000 39999
I hope this is what you want to do
After multiple tries I found that the column had some garbage so used the code below and it worked perfectly. Funny thing is I never encountered the problem on two other data-sets that I imported from excel file.
data2['Code'] = data2['Code'].str.strip()
How do I select columns a and b from df, and save them into a new dataframe df1?
index a b c
1 2 3 4
2 3 4 5
Unsuccessful attempt:
df1 = df['a':'b']
df1 = df.ix[:, 'a':'b']
The column names (which are strings) cannot be sliced in the manner you tried.
Here you have a couple of options. If you know from context which variables you want to slice out, you can just return a view of only those columns by passing a list into the __getitem__ syntax (the []'s).
df1 = df[['a', 'b']]
Alternatively, if it matters to index them numerically and not by their name (say your code should automatically do this without knowing the names of the first two columns) then you can do this instead:
df1 = df.iloc[:, 0:2] # Remember that Python does not slice inclusive of the ending index.
Additionally, you should familiarize yourself with the idea of a view into a Pandas object vs. a copy of that object. The first of the above methods will return a new copy in memory of the desired sub-object (the desired slices).
Sometimes, however, there are indexing conventions in Pandas that don't do this and instead give you a new variable that just refers to the same chunk of memory as the sub-object or slice in the original object. This will happen with the second way of indexing, so you can modify it with the .copy() method to get a regular copy. When this happens, changing what you think is the sliced object can sometimes alter the original object. Always good to be on the look out for this.
df1 = df.iloc[0, 0:2].copy() # To avoid the case where changing df1 also changes df
To use iloc, you need to know the column positions (or indices). As the column positions may change, instead of hard-coding indices, you can use iloc along with get_loc function of columns method of dataframe object to obtain column indices.
{df.columns.get_loc(c): c for idx, c in enumerate(df.columns)}
Now you can use this dictionary to access columns through names and using iloc.
As of version 0.11.0, columns can be sliced in the manner you tried using the .loc indexer:
df.loc[:, 'C':'E']
is equivalent to
df[['C', 'D', 'E']] # or df.loc[:, ['C', 'D', 'E']]
and returns columns C through E.
A demo on a randomly generated DataFrame:
import pandas as pd
import numpy as np
np.random.seed(5)
df = pd.DataFrame(np.random.randint(100, size=(100, 6)),
columns=list('ABCDEF'),
index=['R{}'.format(i) for i in range(100)])
df.head()
Out:
A B C D E F
R0 99 78 61 16 73 8
R1 62 27 30 80 7 76
R2 15 53 80 27 44 77
R3 75 65 47 30 84 86
R4 18 9 41 62 1 82
To get the columns from C to E (note that unlike integer slicing, E is included in the columns):
df.loc[:, 'C':'E']
Out:
C D E
R0 61 16 73
R1 30 80 7
R2 80 27 44
R3 47 30 84
R4 41 62 1
R5 5 58 0
...
The same works for selecting rows based on labels. Get the rows R6 to R10 from those columns:
df.loc['R6':'R10', 'C':'E']
Out:
C D E
R6 51 27 31
R7 83 19 18
R8 11 67 65
R9 78 27 29
R10 7 16 94
.loc also accepts a Boolean array so you can select the columns whose corresponding entry in the array is True. For example, df.columns.isin(list('BCD')) returns array([False, True, True, True, False, False], dtype=bool) - True if the column name is in the list ['B', 'C', 'D']; False, otherwise.
df.loc[:, df.columns.isin(list('BCD'))]
Out:
B C D
R0 78 61 16
R1 27 30 80
R2 53 80 27
R3 65 47 30
R4 9 41 62
R5 78 5 58
...
Assuming your column names (df.columns) are ['index','a','b','c'], then the data you want is in the
third and fourth columns. If you don't know their names when your script runs, you can do this
newdf = df[df.columns[2:4]] # Remember, Python is zero-offset! The "third" entry is at slot two.
As EMS points out in his answer, df.ix slices columns a bit more concisely, but the .columns slicing interface might be more natural, because it uses the vanilla one-dimensional Python list indexing/slicing syntax.
Warning: 'index' is a bad name for a DataFrame column. That same label is also used for the real df.index attribute, an Index array. So your column is returned by df['index'] and the real DataFrame index is returned by df.index. An Index is a special kind of Series optimized for lookup of its elements' values. For df.index it's for looking up rows by their label. That df.columns attribute is also a pd.Index array, for looking up columns by their labels.
In the latest version of Pandas there is an easy way to do exactly this. Column names (which are strings) can be sliced in whatever manner you like.
columns = ['b', 'c']
df1 = pd.DataFrame(df, columns=columns)
In [39]: df
Out[39]:
index a b c
0 1 2 3 4
1 2 3 4 5
In [40]: df1 = df[['b', 'c']]
In [41]: df1
Out[41]:
b c
0 3 4
1 4 5
With Pandas,
wit column names
dataframe[['column1','column2']]
to select by iloc and specific columns with index number:
dataframe.iloc[:,[1,2]]
with loc column names can be used like
dataframe.loc[:,['column1','column2']]
You can use the pandas.DataFrame.filter method to either filter or reorder columns like this:
df1 = df.filter(['a', 'b'])
This is also very useful when you are chaining methods.
You could provide a list of columns to be dropped and return back the DataFrame with only the columns needed using the drop() function on a Pandas DataFrame.
Just saying
colsToDrop = ['a']
df.drop(colsToDrop, axis=1)
would return a DataFrame with just the columns b and c.
The drop method is documented here.
I found this method to be very useful:
# iloc[row slicing, column slicing]
surveys_df.iloc [0:3, 1:4]
More details can be found here.
Starting with 0.21.0, using .loc or [] with a list with one or more missing labels is deprecated in favor of .reindex. So, the answer to your question is:
df1 = df.reindex(columns=['b','c'])
In prior versions, using .loc[list-of-labels] would work as long as at least one of the keys was found (otherwise it would raise a KeyError). This behavior is deprecated and now shows a warning message. The recommended alternative is to use .reindex().
Read more at Indexing and Selecting Data.
You can use Pandas.
I create the DataFrame:
import pandas as pd
df = pd.DataFrame([[1, 2,5], [5,4, 5], [7,7, 8], [7,6,9]],
index=['Jane', 'Peter','Alex','Ann'],
columns=['Test_1', 'Test_2', 'Test_3'])
The DataFrame:
Test_1 Test_2 Test_3
Jane 1 2 5
Peter 5 4 5
Alex 7 7 8
Ann 7 6 9
To select one or more columns by name:
df[['Test_1', 'Test_3']]
Test_1 Test_3
Jane 1 5
Peter 5 5
Alex 7 8
Ann 7 9
You can also use:
df.Test_2
And you get column Test_2:
Jane 2
Peter 4
Alex 7
Ann 6
You can also select columns and rows from these rows using .loc(). This is called "slicing". Notice that I take from column Test_1 to Test_3:
df.loc[:, 'Test_1':'Test_3']
The "Slice" is:
Test_1 Test_2 Test_3
Jane 1 2 5
Peter 5 4 5
Alex 7 7 8
Ann 7 6 9
And if you just want Peter and Ann from columns Test_1 and Test_3:
df.loc[['Peter', 'Ann'], ['Test_1', 'Test_3']]
You get:
Test_1 Test_3
Peter 5 5
Ann 7 9
If you want to get one element by row index and column name, you can do it just like df['b'][0]. It is as simple as you can imagine.
Or you can use df.ix[0,'b'] - mixed usage of index and label.
Note: Since v0.20, ix has been deprecated in favour of loc / iloc.
df[['a', 'b']] # Select all rows of 'a' and 'b'column
df.loc[0:10, ['a', 'b']] # Index 0 to 10 select column 'a' and 'b'
df.loc[0:10, 'a':'b'] # Index 0 to 10 select column 'a' to 'b'
df.iloc[0:10, 3:5] # Index 0 to 10 and column 3 to 5
df.iloc[3, 3:5] # Index 3 of column 3 to 5
Try to use pandas.DataFrame.get (see the documentation):
import pandas as pd
import numpy as np
dates = pd.date_range('20200102', periods=6)
df = pd.DataFrame(np.random.randn(6, 4), index=dates, columns=list('ABCD'))
df.get(['A', 'C'])
One different and easy approach: iterating rows
Using iterows
df1 = pd.DataFrame() # Creating an empty dataframe
for index,i in df.iterrows():
df1.loc[index, 'A'] = df.loc[index, 'A']
df1.loc[index, 'B'] = df.loc[index, 'B']
df1.head()
The different approaches discussed in the previous answers are based on the assumption that either the user knows column indices to drop or subset on, or the user wishes to subset a dataframe using a range of columns (for instance between 'C' : 'E').
pandas.DataFrame.drop() is certainly an option to subset data based on a list of columns defined by user (though you have to be cautious that you always use copy of dataframe and inplace parameters should not be set to True!!)
Another option is to use pandas.columns.difference(), which does a set difference on column names, and returns an index type of array containing desired columns. Following is the solution:
df = pd.DataFrame([[2,3,4], [3,4,5]], columns=['a','b','c'], index=[1,2])
columns_for_differencing = ['a']
df1 = df.copy()[df.columns.difference(columns_for_differencing)]
print(df1)
The output would be:
b c
1 3 4
2 4 5
You can also use df.pop():
>>> df = pd.DataFrame([('falcon', 'bird', 389.0),
... ('parrot', 'bird', 24.0),
... ('lion', 'mammal', 80.5),
... ('monkey', 'mammal', np.nan)],
... columns=('name', 'class', 'max_speed'))
>>> df
name class max_speed
0 falcon bird 389.0
1 parrot bird 24.0
2 lion mammal 80.5
3 monkey mammal
>>> df.pop('class')
0 bird
1 bird
2 mammal
3 mammal
Name: class, dtype: object
>>> df
name max_speed
0 falcon 389.0
1 parrot 24.0
2 lion 80.5
3 monkey NaN
Please use df.pop(c).
I've seen several answers on that, but one remained unclear to me. How would you select those columns of interest?
The answer to that is that if you have them gathered in a list, you can just reference the columns using the list.
Example
print(extracted_features.shape)
print(extracted_features)
(63,)
['f000004' 'f000005' 'f000006' 'f000014' 'f000039' 'f000040' 'f000043'
'f000047' 'f000048' 'f000049' 'f000050' 'f000051' 'f000052' 'f000053'
'f000054' 'f000055' 'f000056' 'f000057' 'f000058' 'f000059' 'f000060'
'f000061' 'f000062' 'f000063' 'f000064' 'f000065' 'f000066' 'f000067'
'f000068' 'f000069' 'f000070' 'f000071' 'f000072' 'f000073' 'f000074'
'f000075' 'f000076' 'f000077' 'f000078' 'f000079' 'f000080' 'f000081'
'f000082' 'f000083' 'f000084' 'f000085' 'f000086' 'f000087' 'f000088'
'f000089' 'f000090' 'f000091' 'f000092' 'f000093' 'f000094' 'f000095'
'f000096' 'f000097' 'f000098' 'f000099' 'f000100' 'f000101' 'f000103']
I have the following list/NumPy array extracted_features, specifying 63 columns. The original dataset has 103 columns, and I would like to extract exactly those, then I would use
dataset[extracted_features]
And you will end up with this
This something you would use quite often in machine learning (more specifically, in feature selection). I would like to discuss other ways too, but I think that has already been covered by other Stack Overflower users.
To exclude some columns you can drop them in the column index. For example:
A B C D
0 1 10 100 1000
1 2 20 200 2000
Select all except two:
df[df.columns.drop(['B', 'D'])]
Output:
A C
0 1 100
1 2 200
You can also use the method truncate to select middle columns:
df.truncate(before='B', after='C', axis=1)
Output:
B C
0 10 100
1 20 200
To select multiple columns, extract and view them thereafter: df is the previously named data frame. Then create a new data frame df1, and select the columns A to D which you want to extract and view.
df1 = pd.DataFrame(data_frame, columns=['Column A', 'Column B', 'Column C', 'Column D'])
df1
All required columns will show up!
def get_slize(dataframe, start_row, end_row, start_col, end_col):
assert len(dataframe) > end_row and start_row >= 0
assert len(dataframe.columns) > end_col and start_col >= 0
list_of_indexes = list(dataframe.columns)[start_col:end_col]
ans = dataframe.iloc[start_row:end_row][list_of_indexes]
return ans
Just use this function
I think this is the easiest way to reach your goal.
import pandas as pd
cols = ['a', 'b']
df1 = pd.DataFrame(df, columns=cols)
df1 = df.iloc[:, 0:2]
I have a pandas dataframe inside a for loop where I change a value in pandas dataframe like this:
df[item].ix[(e1,e2)] = 1
However when I access the df, the values are still unchanged. Do you know where exactly am I going wrong?
Any suggestions?
You are using chained indexing, which usually causes problems. In your code, df[item] returns a series, and then .ix[(e1,e2)] = 1 modifies that series, leaving the original dataframe untouched. You need to modify the original dataframe instead, like this:
import pandas as pd
df = pd.DataFrame({'colA': [5, 6, 1, 2, 3],
'colB': ['a', 'b', 'c', 'd', 'e']})
print df
df.ix[[1, 2], 'colA'] = 111
print df
That code sets rows 1 and 2 of colA to 111, which I believe is the kind of thing you were looking to do. 1 and 2 could be replaced with variables of course.
colA colB
0 5 a
1 6 b
2 1 c
3 2 d
4 3 e
colA colB
0 5 a
1 111 b
2 111 c
3 2 d
4 3 e
For more information on chained indexing, see the documentation:
https://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
Side note: you may also want to rethink your code in general since you mentioned modifying a dataframe in a loop. When using pandas, you usually can and should avoid looping and leverage set-based operations instead. It takes some getting used to, but it's the way to unlock the full power of the library.