Develop Reference

How to replace one of the levels of a MultiIndex dataframe with one of its columns

I have a dataframe such as
multiindex1 = pd.MultiIndex.from_product([['a'], np.arange(3, 8)])
df1 = pd.DataFrame(np.random.randn(5, 3), index=multiindex1)
multiindex2 = pd.MultiIndex.from_product([['s'], np.arange(1, 6)])
df2 = pd.DataFrame(np.random.randn(5, 3), index=multiindex2)
multiindex3 = pd.MultiIndex.from_product([['d'], np.arange(2, 7)])
df3 = pd.DataFrame(np.random.randn(5, 3), index=multiindex3)
df = pd.concat([df1, df2, df3])
df.index.names = ['contract', 'index']
df.columns = ['z', 'x', 'c']
>>>
z x c
contract index
a 3 0.354879 0.206557 0.308081
4 0.822102 -0.425685 1.973288
5 -0.801313 -2.101411 -0.707400
6 -0.740651 -0.564597 -0.975532
7 -0.310679 0.515918 -1.213565
s 1 -0.175135 0.777495 0.100466
2 2.295485 0.381226 -0.242292
3 -0.753414 1.172924 0.679314
4 -0.029526 -0.020714 1.546317
5 0.250066 -1.673020 -0.773842
d 2 -0.602578 -0.761066 -1.117238
3 -0.935758 0.448322 -2.135439
4 0.808704 -0.604837 -0.319351
5 0.321139 0.584896 -0.055951
6 0.041849 -1.660013 -2.157992
Now I want to replace the index of index with the column c. That is to say, I want the result as
z x
contract c
a 0.308081 0.354879 0.206557
1.973288 0.822102 -0.425685
-0.707400 -0.801313 -2.101411
-0.975532 -0.740651 -0.564597
-1.213565 -0.310679 0.515918
s 0.100466 -0.175135 0.777495
-0.242292 2.295485 0.381226
0.679314 -0.753414 1.172924
1.546317 -0.029526 -0.020714
-0.773842 0.250066 -1.673020
d -1.117238 -0.602578 -0.761066
-2.135439 -0.935758 0.448322
-0.319351 0.808704 -0.604837
-0.055951 0.321139 0.584896
-2.157992 0.041849 -1.660013
I implement it in one way
df.reset_index().set_index(['contract', 'c']).drop(['index'], axis=1)
But it seems there are some duplecate steps because I manipulate the indexs for three times. So if there is a more elegent way to achieve that?

Try this
# convert column "c" into an index and remove "index" from index
df.set_index('c', append=True).droplevel('index')
Explanation:
Pandas' set_index method has append argument that controls whether to append columns to existing index or not; setting it True appends column "c" as an index. droplevel method removes index level (can remove column level too but removes index level by default).

Filter Columns from Pandas Dataframe with given list when list elements may or may not be present as column

I have a huge dataframe and I need to filter out the columns from the dataframe if the columns are present in a given list.
For example,
df = pd.DataFrame([[1,2,3,4,5],[6,7,8,9,10]], columns=list('ABCDE'))
This is the dataframe.
A B C D E
0 1 2 3 4 5
1 6 7 8 9 10
I have a list.
fil_lst = ['A', 'D', 'F']
The list may contain column names that are not present in the dataframe. I need only the columns that are present in the dataframe.
I need the resulting dataframe like,
A D
0 1 4
1 6 9
I know it can be done with the help of list comprehension like,
new_df = df[[col for col in fil_lst if col in df.columns]]
But as I have a huge dataframe, it is better if I don't use this computationally expensive process.
Is it possible to vectorize this in any way?

Use Index.isin for test membership in columns and DataFrame.loc for filter by columns, so : mean select all rows and columns by mask:
fil_lst = ['A', 'D', 'F']
df = df.loc[:, df.columns.isin(fil_lst)]
print(df)
A D
0 1 4
1 6 9
Or use Index.intersection:
fil_lst = ['A', 'D', 'F']
df = df[df.columns.intersection(fil_lst)]
print(df)
A D
0 1 4
1 6 9

If you are dealing with large lists, and the focus is on performance more than order of columns, you can use set intersection:
In [2944]: fil_lst = ['A', 'D', 'F']
In [2945]: col_list = df.columns.tolist()
In [2947]: df = df[list(set(col_list) & set(fil_lst))]
In [2947]: df
Out[2947]:
D A
0 4 1
1 9 6
EDIT: If order of columns is important, then do this:
In [2953]: df = df[sorted(set(col_list) & set(fil_lst), key = col_list.index)]
In [2953]: df
Out[2953]:
A D
0 1 4
1 6 9

Accessing an Non Numerical Index in a DataFrame [duplicate]

I'm simply trying to access named pandas columns by an integer.
You can select a row by location using df.ix[3].
But how to select a column by integer?
My dataframe:
df=pandas.DataFrame({'a':np.random.rand(5), 'b':np.random.rand(5)})

Two approaches that come to mind:
>>> df
A B C D
0 0.424634 1.716633 0.282734 2.086944
1 -1.325816 2.056277 2.583704 -0.776403
2 1.457809 -0.407279 -1.560583 -1.316246
3 -0.757134 -1.321025 1.325853 -2.513373
4 1.366180 -1.265185 -2.184617 0.881514
>>> df.iloc[:, 2]
0 0.282734
1 2.583704
2 -1.560583
3 1.325853
4 -2.184617
Name: C
>>> df[df.columns[2]]
0 0.282734
1 2.583704
2 -1.560583
3 1.325853
4 -2.184617
Name: C
Edit: The original answer suggested the use of df.ix[:,2] but this function is now deprecated. Users should switch to df.iloc[:,2].

You can also use df.icol(n) to access a column by integer.
Update: icol is deprecated and the same functionality can be achieved by:
df.iloc[:, n] # to access the column at the nth position

You could use label based using .loc or index based using .iloc method to do column-slicing including column ranges:
In [50]: import pandas as pd
In [51]: import numpy as np
In [52]: df = pd.DataFrame(np.random.rand(4,4), columns = list('abcd'))
In [53]: df
Out[53]:
a b c d
0 0.806811 0.187630 0.978159 0.317261
1 0.738792 0.862661 0.580592 0.010177
2 0.224633 0.342579 0.214512 0.375147
3 0.875262 0.151867 0.071244 0.893735
In [54]: df.loc[:, ["a", "b", "d"]] ### Selective columns based slicing
Out[54]:
a b d
0 0.806811 0.187630 0.317261
1 0.738792 0.862661 0.010177
2 0.224633 0.342579 0.375147
3 0.875262 0.151867 0.893735
In [55]: df.loc[:, "a":"c"] ### Selective label based column ranges slicing
Out[55]:
a b c
0 0.806811 0.187630 0.978159
1 0.738792 0.862661 0.580592
2 0.224633 0.342579 0.214512
3 0.875262 0.151867 0.071244
In [56]: df.iloc[:, 0:3] ### Selective index based column ranges slicing
Out[56]:
a b c
0 0.806811 0.187630 0.978159
1 0.738792 0.862661 0.580592
2 0.224633 0.342579 0.214512
3 0.875262 0.151867 0.071244

You can access multiple columns by passing a list of column indices to dataFrame.ix.
For example:
>>> df = pandas.DataFrame({
'a': np.random.rand(5),
'b': np.random.rand(5),
'c': np.random.rand(5),
'd': np.random.rand(5)
})
>>> df
a b c d
0 0.705718 0.414073 0.007040 0.889579
1 0.198005 0.520747 0.827818 0.366271
2 0.974552 0.667484 0.056246 0.524306
3 0.512126 0.775926 0.837896 0.955200
4 0.793203 0.686405 0.401596 0.544421
>>> df.ix[:,[1,3]]
b d
0 0.414073 0.889579
1 0.520747 0.366271
2 0.667484 0.524306
3 0.775926 0.955200
4 0.686405 0.544421

The method .transpose() converts columns to rows and rows to column, hence you could even write
df.transpose().ix[3]

Most of the people have answered how to take columns starting from an index. But there might be some scenarios where you need to pick columns from in-between or specific index, where you can use the below solution.
Say that you have columns A,B and C. If you need to select only column A and C you can use the below code.
df = df.iloc[:, [0,2]]
where 0,2 specifies that you need to select only 1st and 3rd column.

You can use the method take. For example, to select first and last columns:
df.take([0, -1], axis=1)

reshape a pandas dataframe

suppose a dataframe like this one:
df = pd.DataFrame([[1,2,3,4],[5,6,7,8],[9,10,11,12]], columns = ['A', 'B', 'A1', 'B1'])
I would like to have a dataframe which looks like:
what does not work:
new_rows = int(df.shape[1]/2) * df.shape[0]
new_cols = 2
df.values.reshape(new_rows, new_cols, order='F')
of course I could loop over the data and make a new list of list but there must be a better way. Any ideas ?

The pd.wide_to_long function is built almost exactly for this situation, where you have many of the same variable prefixes that end in a different digit suffix. The only difference here is that your first set of variables don't have a suffix, so you will need to rename your columns first.
The only issue with pd.wide_to_long is that it must have an identification variable, i, unlike melt. reset_index is used to create a this uniquely identifying column, which is dropped later. I think this might get corrected in the future.
df1 = df.rename(columns={'A':'A1', 'B':'B1', 'A1':'A2', 'B1':'B2'}).reset_index()
pd.wide_to_long(df1, stubnames=['A', 'B'], i='index', j='id')\
.reset_index()[['A', 'B', 'id']]
A B id
0 1 2 1
1 5 6 1
2 9 10 1
3 3 4 2
4 7 8 2
5 11 12 2

You can use lreshape, for column id numpy.repeat:
a = [col for col in df.columns if 'A' in col]
b = [col for col in df.columns if 'B' in col]
df1 = pd.lreshape(df, {'A' : a, 'B' : b})
df1['id'] = np.repeat(np.arange(len(df.columns) // 2), len (df.index)) + 1
print (df1)
A B id
0 1 2 1
1 5 6 1
2 9 10 1
3 3 4 2
4 7 8 2
5 11 12 2
EDIT:
lreshape is currently undocumented, but it is possible it might be removed(with pd.wide_to_long too).
Possible solution is merging all 3 functions to one - maybe melt, but now it is not implementated. Maybe in some new version of pandas. Then my answer will be updated.

I solved this in 3 steps:
Make a new dataframe df2 holding only the data you want to be added to the initial dataframe df.
Delete the data from df that will be added below (and that was used to make df2.
Append df2 to df.
Like so:
# step 1: create new dataframe
df2 = df[['A1', 'B1']]
df2.columns = ['A', 'B']
# step 2: delete that data from original
df = df.drop(["A1", "B1"], 1)
# step 3: append
df = df.append(df2, ignore_index=True)
Note how when you do df.append() you need to specify ignore_index=True so the new columns get appended to the index rather than keep their old index.
Your end result should be your original dataframe with the data rearranged like you wanted:
In [16]: df
Out[16]:
A B
0 1 2
1 5 6
2 9 10
3 3 4
4 7 8
5 11 12

Use pd.concat() like so:
#Split into separate tables
df_1 = df[['A', 'B']]
df_2 = df[['A1', 'B1']]
df_2.columns = ['A', 'B'] # Make column names line up
# Add the ID column
df_1 = df_1.assign(id=1)
df_2 = df_2.assign(id=2)
# Concatenate
pd.concat([df_1, df_2])

Merging and sum up several value-counts series in Pandas

I usually use value_counts() to get the number of occurrences of a value. However, I deal now with large database-tables (cannot load it fully into RAM) and query the data in fractions of 1 month.
Is there a way to store the result of value_counts() and merge it with / add it to the next results?
I want to count the number user actions. Assume the following structure of
user-activity logs:
# month 1
id userId actionType
1 1 a
2 1 c
3 2 a
4 3 a
5 3 b
# month 2
id userId actionType
6 1 b
7 1 b
8 2 a
9 3 c
Using value_counts() on those produces:
# month 1
userId
1 2
2 1
3 2
# month 2
userId
1 2
2 1
3 1
Expected output:
# month 1+2
userId
1 4
2 2
3 3
Up until now, I just have found a method using groupby and sum:
# count users actions and remember them in new column
df1['count'] = df1.groupby(['userId'], sort=False)['id'].transform('count')
# delete not necessary columns
df1 = df1[['userId', 'count']]
# delete not necessary rows
df1 = df1.drop_duplicates(subset=['userId'])
# repeat
df2['count'] = df2.groupby(['userId'], sort=False)['id'].transform('count')
df2 = df2[['userId', 'count']]
df2 = df2.drop_duplicates(subset=['userId'])
# merge and sum up
print pd.concat([df1,df2]).groupby(['userId'], sort=False).sum()
What is the pythonic / pandas' way of merging the information of several series' (and dataframes) efficiently?

Let me suggest "add" and specify a fill value of 0. This has an advantage over the previously suggested answer in that it will work when the two Dataframes have non-identical sets of unique keys.
# Create frames
df1 = pd.DataFrame(
{'User_id': ['a', 'a', 'b', 'c', 'c', 'd'], 'a': [1, 1, 2, 3, 3, 5]})
df2 = pd.DataFrame(
{'User_id': ['a', 'a', 'b', 'b', 'c', 'c', 'c'], 'a': [1, 1, 2, 2, 3, 3, 4]})
Now add the the two sets of values_counts(). The fill_value argument will handle any NaN values that would arise, in this example, the 'd' that appears in df1, but not df2.
a = df1.User_id.value_counts()
b = df2.User_id.value_counts()
a.add(b,fill_value=0)

You can sum the series generated by the value_counts method directly:
#create frames
df= pd.DataFrame({'User_id': ['a','a','b','c','c'],'a':[1,1,2,3,3]})
df1= pd.DataFrame({'User_id': ['a','a','b','b','c','c','c'],'a':[1,1,2,2,3,3,4]})
sum the series:
df.User_id.value_counts() + df1.User_id.value_counts()
output:
a 4
b 3
c 5
dtype: int64

This is know as "Split-Apply-Combine". It is done in 1 line and 3-4 clicks, using a lambda function as follows.
1️⃣ paste this into your code:
df['total_for_this_label'] = df.groupby('label', as_index=False)['label'].transform(lambda x: x.count())
2️⃣ replace 3x label with the name of the column whose values you are counting (case-sensitive)
3️⃣ print df.head() to check it's worked correctly