Related
I would like to go through every row (entry) in my df and remove every entry that has the value of " " (which yes is an empty string).
So if my data set is:
Name Gender Age
Jack 5
Anna F 6
Carl M 7
Jake M 7
Therefore Jack would be removed from the dataset.
On another note, I would also like to remove entries that has the value "Unspecified" and "Undetermined" as well.
Eg:
Name Gender Age Address
Jack 5 *address*
Anna F 6 *address*
Carl M 7 Undetermined
Jake M 7 Unspecified
Now,
Jack will be removed due to empty field.
Carl will be removed due to the value Undetermined present in a column.
Jake will be removed due to the value Unspecified present in a column.
For now, this has been my approach but I keep getting a TypeError.
list = []
for i in df.columns:
if df[i] == "":
# everytime there is an empty string, add 1 to list
list.append(1)
# count list to see how many entries there are with empty string
len(list)
Please help me with this. I would prefer a for loop being used due to there being about 22 columns and 9000+ rows in my actual dataset.
Note - I do understand that there are other questions asked like this, its just that none of them apply to my situation, meaning that most of them are only useful for a few columns and I do not wish to hardcode all 22 columns.
Edit - Thank you for all your feedbacks, you all have been incredibly helpful.
To delete a row based on a condition use the following:
df = df.drop(df[condition].index)
For example:
df = df.drop(df[Age==5].index) , will drop the row where the Age is 5.
I've come across a post regarding the same dating back to 2017, it should help you understand it more clearer.
Regarding question 2, here's how to remove rows with the specified values in a given column:
df = df[~df["Address"].isin(("Undetermined", "Unspecified"))]
Let's assume we have a Pandas DataFrame object df.
To remove every row given your conditions, simply do:
df = df[df.Gender == " " or df.df.Age == " " or df.Address in [" ", "Undetermined", "Unspecified"]]
If the unspecified fields are NaN, you can also do:
df = df.dropna(how="any", axis = 0)
Answer from #ThatCSFresher or #Bence will help you out in removing rows based on single column... Which is great!
However, I think there are multiple condition in your query needed to check across multiple columns at once in a loop. So, probably apply-lambda can do the job; Try the following code;
df = pd.DataFrame({"Name":["Jack","Anna","Carl","Jake"],
"Gender":["","F","M","M"],
"Age":[5,6,7,7],
"Address":["address","address","Undetermined","Unspecified"]})
df["Noise_Tag"] = df.apply(lambda x: "Noise" if ("" in list(x)) or ("Undetermined" in list(x)) or ("Unspecified" in list(x)) else "No Noise",axis=1)
df1 = df[df["Noise_Tag"] == "No Noise"]
del df1["Noise_Tag"]
# Output of df;
Name Gender Age Address Noise_Tag
0 Jack 5 address Noise
1 Anna F 6 address No Noise
2 Carl M 7 Undetermined Noise
3 Jake M 7 Unspecified Noise
# Output of df1;
Name Gender Age Address
1 Anna F 6 address
Well, OP actually wants to delete any column with "empty" string.
df = df[~(df=="").any(axis=1)] # deletes all rows that have empty string in any column.
If you want to delete specifically for address column, then you can just delete using
df = df[~df["Address"].isin(("Undetermined", "Unspecified"))]
Or if any column with Undetermined or Unspecified, try similar as the first solution in my post, just by replacing the empty string with Undertermined or Unspecified.
df = df[~((df=="Undetermined") | (df=="Unspecified")).any(axis=1)]
You can build masks and then filter the df according to it:
m1 = df.eq('').any(axis=1)
# m1 is True if any cell in a row has an empty string
m2 = df['Address'].isin(['Undetermined', 'Unspecified'])
# m2 is True if a row has one of the values in the list in column 'Address'
out = df[~m1 & ~m2] # invert both condition and get the desired output
print(out)
Output:
Name Gender Age Address
1 Anna F 6 *address*
Used Input:
df = pd.DataFrame({'Name': ['Jack', 'Anna', 'Carl', 'Jake'],
'Gender': ['', 'F', 'M', 'M'],
'Age': [5, 6, 7, 7],
'Address': ['*address*', '*address*', 'Undetermined', 'Unspecified']}
)
using lambda fun
Code:
df[df.apply(lambda x: False if (x.Address in ['Undetermined', 'Unspecified'] or '' in list(x)) else True, axis=1)]
Output:
Name Gender Age Address
1 Anna F 6 *add
def stack_plot(data, xtick, col2='project_is_approved', col3='total'):
ind = np.arange(data.shape[0])
plt.figure(figsize=(20,5))
p1 = plt.bar(ind, data[col3].values)
p2 = plt.bar(ind, data[col2].values)
plt.ylabel('Projects')
plt.title('Number of projects aproved vs rejected')
plt.xticks(ind, list(data[xtick].values))
plt.legend((p1[0], p2[0]), ('total', 'accepted'))
plt.show()
def univariate_barplots(data, col1, col2='project_is_approved', top=False):
# Count number of zeros in dataframe python: https://stackoverflow.com/a/51540521/4084039
temp = pd.DataFrame(project_data.groupby(col1)[col2].agg(lambda x: x.eq(1).sum())).reset_index()
# Pandas dataframe grouby count: https://stackoverflow.com/a/19385591/4084039
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'Avg':'mean'})).reset_index()['Avg']
temp.sort_values(by=['total'],inplace=True, ascending=False)
if top:
temp = temp[0:top]
stack_plot(temp, xtick=col1, col2=col2, col3='total')
print(temp.head(5))
print("="*50)
print(temp.tail(5))
univariate_barplots(project_data, 'school_state', 'project_is_approved', False)
Error:
SpecificationError Traceback (most recent call last)
<ipython-input-21-2cace8f16608> in <module>()
----> 1 univariate_barplots(project_data, 'school_state', 'project_is_approved', False)
<ipython-input-20-856fcc83737b> in univariate_barplots(data, col1, col2, top)
4
5 # Pandas dataframe grouby count: https://stackoverflow.com/a/19385591/4084039
----> 6 temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
7 print (temp['total'].head(2))
8 temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'Avg':'mean'})).reset_index()['Avg']
~\AppData\Roaming\Python\Python36\site-packages\pandas\core\groupby\generic.py in aggregate(self, func, *args, **kwargs)
251 # but not the class list / tuple itself.
252 func = _maybe_mangle_lambdas(func)
--> 253 ret = self._aggregate_multiple_funcs(func)
254 if relabeling:
255 ret.columns = columns
~\AppData\Roaming\Python\Python36\site-packages\pandas\core\groupby\generic.py in _aggregate_multiple_funcs(self, arg)
292 # GH 15931
293 if isinstance(self._selected_obj, Series):
--> 294 raise SpecificationError("nested renamer is not supported")
295
296 columns = list(arg.keys())
SpecificationError: **nested renamer is not supported**
change
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'Avg':'mean'})).reset_index()['Avg']
to
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg(total='count')).reset_index()['total']
temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg(Avg='mean')).reset_index()['Avg']
reason: in new pandas version named aggregation is the recommended replacement for the deprecated “dict-of-dicts” approach to naming the output of column-specific aggregations (Deprecate groupby.agg() with a dictionary when renaming).
source: https://pandas.pydata.org/pandas-docs/stable/whatsnew/v0.25.0.html
This error also happens if a column specified in the aggregation function dict does not exist in the dataframe:
In [190]: group = pd.DataFrame([[1, 2]], columns=['A', 'B']).groupby('A')
In [195]: group.agg({'B': 'mean'})
Out[195]:
B
A
1 2
In [196]: group.agg({'B': 'mean', 'non-existing-column': 'mean'})
...
SpecificationError: nested renamer is not supported
I found the way: Instead of going like
g2 = df.groupby(["Description","CustomerID"],as_index=False).agg({'Quantity':{"maxQ":np.max,"minQ":np.min,"meanQ":np.mean}})
g2.columns = ["Description","CustomerID","maxQ","minQ",'meanQ']
Do as follows:
g2 = df.groupby(["Description","CustomerID"],as_index=False).agg({'Quantity':{np.max,np.min,np.mean}})
g2.columns = ["Description","CustomerID","maxQ","minQ",'meanQ']
I had the same error and this is how I resolved it!
Do you get the same error if you change
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
to
temp['total'] = project_data.groupby(col1)[col2].agg(total=('total','count')).reset_index()['total']
Instead of using .agg({'total':'count'})), you can pass name with the function as a list of tuple like .agg([('total', 'count')])and use the same for Avg also. Hope it would work.
I have got the similar issue as #akshay jindal, but I check the documentation as suggested by #artikay Khanna, the problem solved, some functions has been adjusted, the old is deprecated. Here is the code warning provided per last time execute.
/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:1: FutureWarning: using a dict on a Series for aggregation
is deprecated and will be removed in a future version. Use named aggregation instead.
>>> grouper.agg(name_1=func_1, name_2=func_2)
"""Entry point for launching an IPython kernel.
Therefore, I will suggest try
grouper.agg(name_1=func_1, name_2=func_2)
Hope this will help
Not a very elegant solution but this one works. As renaming the column is deprecated with the way you are doing. But there is work around. Create a temporary variable 'approved' , store the col2 in it. Because when you apply agg function , the original column values will change with column name. You can preserve the column name but then values in those column will change. So in order to preserve the original dataframe and to have two new columns with desired names, you can use the following code.
approved = temp[col2]
temp = pd.DataFrame(project_data.groupby(col1)[col2].agg([('Avg','mean'),('total','count')]).reset_index())
temp[col2] = approved
P.S : Seems like an assignment of AAIC, I am working on same :)
Sometimes it's convenient to keep an aggdict of how each column should be transformed under aggregation that will work with different column sets and different group by columns. You can do this with the new syntax fairly easily by unpacking the dict with **. Here's a minimal working example for simple data.
dfx=pd.DataFrame(columns=["A","B","C"],data=np.random.randint(0,5,size=(10,3)))
#dfx
#
# A B C
#0 4 4 1
#1 2 4 4
#2 1 3 3
#3 2 4 3
#4 1 2 1
#5 0 4 2
#6 2 3 4
#7 1 0 2
#8 2 1 4
#9 3 0 3
Maybe when you agg you want the first "A", the last "B", the mean "C" and sometimes your pipeline has a "D" (but not this time) that you also want the mean of.
aggdict = {"A":lambda x: x.iloc[0], "B": lambda x: x.iloc[-1], "C" : "mean" , "D":lambda x: "mean"}
You can build a simple dict like the old days and then unpack it with ** filtering on the relevant keys:
gb_col="C"
gbc = dfx.groupby(gb_col).agg(**{k:(k,v) for k,v in aggdict.items() if k in dfx.columns and k != gb_col})
# A B
#C
#1 4 2
#2 0 0
#3 1 4
#4 2 3
And then you can slice and dice how you want with the same syntax:
mygb = lambda gb_col: dfx.groupby(gb_col).agg(**{k:(k,v) for k,v in aggdict.items() if k in dfx.columns and k != gb_col})
allgb = [mygb(c) for c in dfx.columns]
I have tried alll the solutions and turned out to be the error with the name. If your column name has some inbuilt keywords such as "in", "is",etc., It is throwing error. In my case, My column name is "Points in Polygon" and I have resolved the issue by renaming the column to "Points"
#Rishi's solution worked for me. The original name of the column in my dataframe was net_value_budgeted_rate, which was essentially dollar value of the sale. I changed it to dollars and it worked.
Info = pd.DataFrame(df.groupby("school_state").agg(Approved=("project_is_approved",lambda x: x.eq(1).sum()),Total=("project_is_approved","count"),Avg=("project_is_approved","mean"))).reset_index().sort_values(by=["Total"],ascending=False).head()
You can break this into individual commands for better readability.
def stack_plot(data, xtick, col2='project_is_approved', col3='total'):
ind = np.arange(data.shape[0])
plt.figure(figsize=(20,5))
p1 = plt.bar(ind, data[col3].values)
p2 = plt.bar(ind, data[col2].values)
plt.ylabel('Projects')
plt.title('Number of projects aproved vs rejected')
plt.xticks(ind, list(data[xtick].values))
plt.legend((p1[0], p2[0]), ('total', 'accepted'))
plt.show()
def univariate_barplots(data, col1, col2='project_is_approved', top=False):
# Count number of zeros in dataframe python: https://stackoverflow.com/a/51540521/4084039
temp = pd.DataFrame(project_data.groupby(col1)[col2].agg(lambda x: x.eq(1).sum())).reset_index()
# Pandas dataframe grouby count: https://stackoverflow.com/a/19385591/4084039
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'Avg':'mean'})).reset_index()['Avg']
temp.sort_values(by=['total'],inplace=True, ascending=False)
if top:
temp = temp[0:top]
stack_plot(temp, xtick=col1, col2=col2, col3='total')
print(temp.head(5))
print("="*50)
print(temp.tail(5))
univariate_barplots(project_data, 'school_state', 'project_is_approved', False)
Error:
SpecificationError Traceback (most recent call last)
<ipython-input-21-2cace8f16608> in <module>()
----> 1 univariate_barplots(project_data, 'school_state', 'project_is_approved', False)
<ipython-input-20-856fcc83737b> in univariate_barplots(data, col1, col2, top)
4
5 # Pandas dataframe grouby count: https://stackoverflow.com/a/19385591/4084039
----> 6 temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
7 print (temp['total'].head(2))
8 temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'Avg':'mean'})).reset_index()['Avg']
~\AppData\Roaming\Python\Python36\site-packages\pandas\core\groupby\generic.py in aggregate(self, func, *args, **kwargs)
251 # but not the class list / tuple itself.
252 func = _maybe_mangle_lambdas(func)
--> 253 ret = self._aggregate_multiple_funcs(func)
254 if relabeling:
255 ret.columns = columns
~\AppData\Roaming\Python\Python36\site-packages\pandas\core\groupby\generic.py in _aggregate_multiple_funcs(self, arg)
292 # GH 15931
293 if isinstance(self._selected_obj, Series):
--> 294 raise SpecificationError("nested renamer is not supported")
295
296 columns = list(arg.keys())
SpecificationError: **nested renamer is not supported**
change
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'Avg':'mean'})).reset_index()['Avg']
to
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg(total='count')).reset_index()['total']
temp['Avg'] = pd.DataFrame(project_data.groupby(col1)[col2].agg(Avg='mean')).reset_index()['Avg']
reason: in new pandas version named aggregation is the recommended replacement for the deprecated “dict-of-dicts” approach to naming the output of column-specific aggregations (Deprecate groupby.agg() with a dictionary when renaming).
source: https://pandas.pydata.org/pandas-docs/stable/whatsnew/v0.25.0.html
This error also happens if a column specified in the aggregation function dict does not exist in the dataframe:
In [190]: group = pd.DataFrame([[1, 2]], columns=['A', 'B']).groupby('A')
In [195]: group.agg({'B': 'mean'})
Out[195]:
B
A
1 2
In [196]: group.agg({'B': 'mean', 'non-existing-column': 'mean'})
...
SpecificationError: nested renamer is not supported
I found the way: Instead of going like
g2 = df.groupby(["Description","CustomerID"],as_index=False).agg({'Quantity':{"maxQ":np.max,"minQ":np.min,"meanQ":np.mean}})
g2.columns = ["Description","CustomerID","maxQ","minQ",'meanQ']
Do as follows:
g2 = df.groupby(["Description","CustomerID"],as_index=False).agg({'Quantity':{np.max,np.min,np.mean}})
g2.columns = ["Description","CustomerID","maxQ","minQ",'meanQ']
I had the same error and this is how I resolved it!
Do you get the same error if you change
temp['total'] = pd.DataFrame(project_data.groupby(col1)[col2].agg({'total':'count'})).reset_index()['total']
to
temp['total'] = project_data.groupby(col1)[col2].agg(total=('total','count')).reset_index()['total']
Instead of using .agg({'total':'count'})), you can pass name with the function as a list of tuple like .agg([('total', 'count')])and use the same for Avg also. Hope it would work.
I have got the similar issue as #akshay jindal, but I check the documentation as suggested by #artikay Khanna, the problem solved, some functions has been adjusted, the old is deprecated. Here is the code warning provided per last time execute.
/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:1: FutureWarning: using a dict on a Series for aggregation
is deprecated and will be removed in a future version. Use named aggregation instead.
>>> grouper.agg(name_1=func_1, name_2=func_2)
"""Entry point for launching an IPython kernel.
Therefore, I will suggest try
grouper.agg(name_1=func_1, name_2=func_2)
Hope this will help
Not a very elegant solution but this one works. As renaming the column is deprecated with the way you are doing. But there is work around. Create a temporary variable 'approved' , store the col2 in it. Because when you apply agg function , the original column values will change with column name. You can preserve the column name but then values in those column will change. So in order to preserve the original dataframe and to have two new columns with desired names, you can use the following code.
approved = temp[col2]
temp = pd.DataFrame(project_data.groupby(col1)[col2].agg([('Avg','mean'),('total','count')]).reset_index())
temp[col2] = approved
P.S : Seems like an assignment of AAIC, I am working on same :)
Sometimes it's convenient to keep an aggdict of how each column should be transformed under aggregation that will work with different column sets and different group by columns. You can do this with the new syntax fairly easily by unpacking the dict with **. Here's a minimal working example for simple data.
dfx=pd.DataFrame(columns=["A","B","C"],data=np.random.randint(0,5,size=(10,3)))
#dfx
#
# A B C
#0 4 4 1
#1 2 4 4
#2 1 3 3
#3 2 4 3
#4 1 2 1
#5 0 4 2
#6 2 3 4
#7 1 0 2
#8 2 1 4
#9 3 0 3
Maybe when you agg you want the first "A", the last "B", the mean "C" and sometimes your pipeline has a "D" (but not this time) that you also want the mean of.
aggdict = {"A":lambda x: x.iloc[0], "B": lambda x: x.iloc[-1], "C" : "mean" , "D":lambda x: "mean"}
You can build a simple dict like the old days and then unpack it with ** filtering on the relevant keys:
gb_col="C"
gbc = dfx.groupby(gb_col).agg(**{k:(k,v) for k,v in aggdict.items() if k in dfx.columns and k != gb_col})
# A B
#C
#1 4 2
#2 0 0
#3 1 4
#4 2 3
And then you can slice and dice how you want with the same syntax:
mygb = lambda gb_col: dfx.groupby(gb_col).agg(**{k:(k,v) for k,v in aggdict.items() if k in dfx.columns and k != gb_col})
allgb = [mygb(c) for c in dfx.columns]
I have tried alll the solutions and turned out to be the error with the name. If your column name has some inbuilt keywords such as "in", "is",etc., It is throwing error. In my case, My column name is "Points in Polygon" and I have resolved the issue by renaming the column to "Points"
#Rishi's solution worked for me. The original name of the column in my dataframe was net_value_budgeted_rate, which was essentially dollar value of the sale. I changed it to dollars and it worked.
Info = pd.DataFrame(df.groupby("school_state").agg(Approved=("project_is_approved",lambda x: x.eq(1).sum()),Total=("project_is_approved","count"),Avg=("project_is_approved","mean"))).reset_index().sort_values(by=["Total"],ascending=False).head()
You can break this into individual commands for better readability.
EDIT: I have stripped down the file to the bits that are problematic
raw_data = {"link":
['https://www.otodom.pl/oferta/mieszkanie-w-spokojnej-okolicy-gdansk-lostowice-ID43FLJ.html#cda8700ef5',
'https://www.otodom.pl/oferta/mieszkanie-w-spokojnej-okolicy-gdansk-lostowice-ID43FLH.html#cda8700ef5',
'https://www.otodom.pl/oferta/mieszkanie-w-spokojnej-okolicy-gdansk-lostowice-ID43FLj.html#cda8700ef5',
'https://www.otodom.pl/oferta/mieszkanie-w-spokojnej-okolicy-gdansk-lostowice-ID43FLh.html#cda8700ef5',
'https://www.otodom.pl/oferta/zielony-widok-mieszkanie-3m04-ID43EWU.html#9dca9667c3',
'https://www.otodom.pl/oferta/zielony-widok-mieszkanie-3m04-ID43EWu.html#9dca9667c3',
'https://www.otodom.pl/oferta/nowoczesne-osiedle-gotowe-do-konca-roku-bazantow-ID43vQM.html#af24036d28',
'https://www.otodom.pl/oferta/nowoczesne-osiedle-gotowe-do-konca-roku-bazantow-ID43vQJ.html#af24036d28',
'https://www.otodom.pl/oferta/nowoczesne-osiedle-gotowe-do-konca-roku-bazantow-ID43vQm.html#af24036d28',
'https://www.otodom.pl/oferta/nowoczesne-osiedle-gotowe-do-konca-roku-bazantow-ID43vQj.html#af24036d28',
'https://www.otodom.pl/oferta/mieszkanie-56-m-warszawa-ID43sWY.html#2d0084b7ea',
'https://www.otodom.pl/oferta/mieszkanie-56-m-warszawa-ID43sWy.html#2d0084b7ea',
'https://www.otodom.pl/oferta/idealny-2pok-apartament-0-pcc-widok-na-park-ID43q4X.html#64f19d3152',
'https://www.otodom.pl/oferta/idealny-2pok-apartament-0-pcc-widok-na-park-ID43q4x.html#64f19d3152']}
df = pd.DataFrame(raw_data, columns = ["link"])
#duplicate check #1
a = print(df.iloc[12][0])
b = print(df.iloc[13][0])
if a == b:
print("equal")
#duplicate check #2
df.duplicated()
For the first test I get the following output implying there is a duplicate
https://www.otodom.pl/oferta/idealny-2pok-apartament-0-pcc-widok-na-park-ID43q4X.html#64f19d3152
https://www.otodom.pl/oferta/idealny-2pok-apartament-0-pcc-widok-na-park-ID43q4x.html#64f19d3152
equal
For the second test it seems there are no duplicates
0 False
1 False
2 False
3 False
4 False
5 False
6 False
7 False
8 False
9 False
10 False
11 False
12 False
13 False
dtype: bool
Original post:
Trying to identify duplicate values from the "Link" column of attached file:
data file
import pandas as pd
data = pd.read_csv(r"...\consolidated.csv", sep=",")
df = pd.DataFrame(data)
del df['Unnamed: 0']
duplicate_rows = df[df.duplicated(["Link"], keep="first")]
pd.DataFrame(duplicate_rows)
#a = print(df.iloc[42657][15])
#b = print(df.iloc[42676][15])
#if a == b:
# print("equal")
Used the code above, but the answer I keep getting is that there are no duplicates. Checked it through Excel and there should be seven duplicate instances. Even selected specific cells to do a quick check (the part marked with #s), and the values have been identified as equal. Yet duplicated does not capture them
I have been scratching my head for a good hour, and still no idea what I'm missing - help appreciated!
I had the same problem and converting the columns of the dataframe to "str" helped.
eg.
df['link'] = df['link'].astype(str)
duplicate_rows = df[df.duplicated(["link"], keep="first")]
First, you don't need df = pd.DataFrame(data), as data = pd.read_csv(r"...\consolidated.csv", sep=",") already returns a Dataframe.
As for the deletion of duplicates, check the drop_duplicates method in the Documentation
Hope this helps.
I have a few functions that make new columns in a pandas dataframe, as a function of existing columns in the dataframe. I have two different scenarios that occur here: (1) the dataframe is NOT multiIndex and has a set of columns, say [a,b] and (2) the dataframe is multiIndex and now has the same set of columns headers repeated N times, say [(a,1),(b,1),(a,2),(b,2)....(a,N),(n,N)].
I've been making the aforementioned functions in the style shown below:
def f(df):
if multiindex(df):
for s df[a].columns:
df[c,s] = someFunction(df[a,s], df[b,s])
else:
df[c] = someFunction(df[a], df[b])
Is there another way to do this, without having these if-multi-index/else statement everywhere and duplicating the someFunction code? I'd prefer NOT to split the multi indexed frame into N smaller dataframes (I often need to filter data or do things and keep the rows consistent across all the 1,2,...N frames, and keeping them together in one frame seems the to be the best way to do that).
you may still have to test if columns is a MultiIndex but this should be cleaner and more efficient. Caveat, this will not work if your function utilizes summary statistics on the column. For example, if someFunction divides by the the average of column 'a'.
Solution
def someFunction(a, b):
return a + b
def f(df):
df = df.copy()
ismi = isinstance(df.columns, pd.MultiIndex)
if ismi:
df = df.stack()
df['c'] = someFunction(df['a'], df['a'])
if ismi:
df = df.unstack()
return df
Setup
import pandas as pd
import numpy as np
setup_tuples = []
for c in ['a', 'b']:
for i in ['one', 'two', 'three']:
setup_tuples.append((c, i))
columns = pd.MultiIndex.from_tuples(setup_tuples)
rand_array = np.random.rand(10, len(setup_tuples))
df = pd.DataFrame(rand_array, columns=columns)
df looks like this
a b
one two three one two three
0 0.282834 0.490313 0.201300 0.140157 0.467710 0.352555
1 0.838527 0.707131 0.763369 0.265170 0.452397 0.968125
2 0.822786 0.785226 0.434637 0.146397 0.056220 0.003197
3 0.314795 0.414096 0.230474 0.595133 0.060608 0.900934
4 0.334733 0.118689 0.054299 0.237786 0.658538 0.057256
5 0.993753 0.552942 0.665615 0.336948 0.788817 0.320329
6 0.310809 0.199921 0.158675 0.059406 0.801491 0.134779
7 0.971043 0.183953 0.723950 0.909778 0.103679 0.695661
8 0.755384 0.728327 0.029720 0.408389 0.808295 0.677195
9 0.276158 0.978232 0.623972 0.897015 0.253178 0.093772
I constructed df to have MultiIndex columns. What I'd do is use the .stack() method to push the second level of the column index to be the second level of the row index.
df.stack() looks like this
a b
0 one 0.282834 0.140157
three 0.201300 0.352555
two 0.490313 0.467710
1 one 0.838527 0.265170
three 0.763369 0.968125
two 0.707131 0.452397
2 one 0.822786 0.146397
three 0.434637 0.003197
two 0.785226 0.056220
3 one 0.314795 0.595133
three 0.230474 0.900934
two 0.414096 0.060608
4 one 0.334733 0.237786
three 0.054299 0.057256
two 0.118689 0.658538
5 one 0.993753 0.336948
three 0.665615 0.320329
two 0.552942 0.788817
6 one 0.310809 0.059406
three 0.158675 0.134779
two 0.199921 0.801491
7 one 0.971043 0.909778
three 0.723950 0.695661
two 0.183953 0.103679
8 one 0.755384 0.408389
three 0.029720 0.677195
two 0.728327 0.808295
9 one 0.276158 0.897015
three 0.623972 0.093772
two 0.978232 0.253178
Now you can operate on df.stack() as if the columns were not a MultiIndex
Demonstration
print f(df)
will give you what you want
a b c \
one three two one three two one
0 0.282834 0.201300 0.490313 0.140157 0.352555 0.467710 0.565667
1 0.838527 0.763369 0.707131 0.265170 0.968125 0.452397 1.677055
2 0.822786 0.434637 0.785226 0.146397 0.003197 0.056220 1.645572
3 0.314795 0.230474 0.414096 0.595133 0.900934 0.060608 0.629591
4 0.334733 0.054299 0.118689 0.237786 0.057256 0.658538 0.669465
5 0.993753 0.665615 0.552942 0.336948 0.320329 0.788817 1.987507
6 0.310809 0.158675 0.199921 0.059406 0.134779 0.801491 0.621618
7 0.971043 0.723950 0.183953 0.909778 0.695661 0.103679 1.942086
8 0.755384 0.029720 0.728327 0.408389 0.677195 0.808295 1.510767
9 0.276158 0.623972 0.978232 0.897015 0.093772 0.253178 0.552317
three two
0 0.402600 0.980626
1 1.526739 1.414262
2 0.869273 1.570453
3 0.460948 0.828193
4 0.108599 0.237377
5 1.331230 1.105884
6 0.317349 0.399843
7 1.447900 0.367907
8 0.059439 1.456654
9 1.247944 1.956464