Having columns with subscripts/indices in pandas - python

Let's say that I have som data from a file where some columns are "of the same kind", only of different subscripts of some mathematical variable, say x:
n A B C x[0] x[1] x[2]
0 1 2 3 4 5 6
1 2 3 4 5 6 7
Is there some way I can load this into a pandas dataframe df and somehow treat the three x-columns as an indexable, array-like entity (I'm new to pandas)? I believe it would be convenient, because I could do operations on the data-series contained in x such as sum(df.x).
Kind regards.
EDIT:
Admittedly, my original post was not clear enough. I'm not just interested in getting the sum of three columns. That was just an example. I'm looking for a generally applicable abstraction that I hope is built into pandas.
I'd like to have multiple columns accessible through (sub-)indices of one entity, e.g. df.x[0], such that I (or any other user of the data) can do whichever operation he/she wants (sum/max/min/avg/standard deviation, you name it). You can consider the x's as an ensamble of time-dependent measurements if you like.
Kind regards.

Consider, you define your dataframe like this
df = pd.DataFrame([[1, 2, 3, 4, 5, 6],
[2, 3, 4, 5, 6, 7]], columns=['A', 'B', 'C', 'x0', 'x1', 'x2'])
Then with
x = ['x0', 'x1', 'x2']
You use the following notation allowing a quite general definition of x
>>> df[x].sum(axis=1)
0 15
1 18
dtype: int64

Look of column which starts with 'x' and perform operations you need
column_num=[col for col in df.columns if col.startswith('x')]
df[column_num].sum(axis=1)

I'll give you another answer which will defer from you initial data structure in exchange for addressing the values of the dataframe by df.x[0] etc.
Consider you have defined your dataframe like this
>>> dv = pd.DataFrame(np.random.randint(10, size=20),
index=pd.MultiIndex.from_product([range(4), range(5)]), columns=['x'])
>>> dv
x
0 0 8
1 3
2 4
3 6
4 1
1 0 8
1 9
2 1
3 8
4 8
[...]
Then you can exactly do this
dv.x[1]
0 8
1 9
2 1
3 8
4 8
Name: x, dtype: int64
which is your desired notation. Requires some changes to your initial set-up but will give you exactly what you want.

Related

Check if many columns of a data frame are exactly the same

I am developing a clinical bioinformatic application and the input this application gets is a data frame that looks like this
df = pd.DataFrame({'store': ['Blank_A09', 'Control_4p','13_MEG3','04_GRB10','02_PLAGL1','Control_21q','01_PLAGL1','11_KCNQ10T1','16_SNRPN','09_H19','Control_6p','06_MEST'],
'quarter': [1, 1, 2, 2, 1, 1, 2, 2,2,2,2,2],
'employee': ['Blank_A09', 'Control_4p','13_MEG3','04_GRB10','02_PLAGL1','Control_21q','01_PLAGL1','11_KCNQ10T1','16_SNRPN','09_H19','Control_6p','06_MEST'],
'foo': [1, 1, 2, 2, 1, 1, 9, 2,2,4,2,2],
'columnX': ['Blank_A09', 'Control_4p','13_MEG3','04_GRB10','02_PLAGL1','Control_21q','01_PLAGL1','11_KCNQ10T1','16_SNRPN','09_H19','Control_6p','06_MEST']})
print(df)
store quarter employee foo columnX
0 Blank_A09 1 Blank_A09 1 Blank_A09
1 Control_4p 1 Control_4p 1 Control_4p
2 13_MEG3 2 13_MEG3 2 13_MEG3
3 04_GRB10 2 04_GRB10 2 04_GRB10
4 02_PLAGL1 1 02_PLAGL1 1 02_PLAGL1
5 Control_21q 1 Control_21q 1 Control_21q
6 01_PLAGL1 2 01_PLAGL1 9 01_PLAGL1
7 11_KCNQ10T1 2 11_KCNQ10T1 2 11_KCNQ10T1
8 16_SNRPN 2 16_SNRPN 2 16_SNRPN
9 09_H19 2 09_H19 4 09_H19
10 Control_6p 2 Control_6p 2 Control_6p
11 06_MEST 2 06_MEST 2 06_MEST
This is a minimal reproducible example, but the real one has an uncertain number of columns in which the first, the third the 5th, the 7th, etc. "should" be exactly the same.
And this is what I want to check. I want to ensure that these columns have their values in the same order.
I know how to check if 2 columns are exactly the same but I don't know how to expand this checking across all data frame.
EDIT:
The name of the columns change, in my example, they are just two examples.
Refer here How to check if 3 columns are same and add a new column with the value if the values are same?
Here is a code that would check if more columns are the same and returns the index of rows which are the same
arr = df[['quarter','foo_test','foo']].values #You can add as many columns as you wish
np.where((arr == arr[:, [0]]).all(axis=1))
You need to tweak it for your usage
Edit
columns_to_check = [x for x in range(1, len(df.columns), 2)]
arr = df.iloc[:, columns_to_check].values
If you want an efficient method you can hash the Series using pandas.util.hash_pandas_object, making the operation O(n):
pd.util.hash_pandas_object(df.T, index=False)
We clearly see that store/employee/columnX have the same hash:
store 18266754969677227875
quarter 11367719614658692759
employee 18266754969677227875
foo 92544834319824418
columnX 18266754969677227875
dtype: uint64
You can further use groupby to identify the identical values:
df.columns.groupby(pd.util.hash_pandas_object(df.T, index=False))
output:
{ 92544834319824418: ['foo'],
11367719614658692759: ['quarter'],
18266754969677227875: ['store', 'employee', 'columnX']}

Sum of a groupby dataframe not equal to the sum of a dataframe [duplicate]

I have a DataFrame with many missing values in columns which I wish to groupby:
import pandas as pd
import numpy as np
df = pd.DataFrame({'a': ['1', '2', '3'], 'b': ['4', np.NaN, '6']})
In [4]: df.groupby('b').groups
Out[4]: {'4': [0], '6': [2]}
see that Pandas has dropped the rows with NaN target values. (I want to include these rows!)
Since I need many such operations (many cols have missing values), and use more complicated functions than just medians (typically random forests), I want to avoid writing too complicated pieces of code.
Any suggestions? Should I write a function for this or is there a simple solution?
pandas >= 1.1
From pandas 1.1 you have better control over this behavior, NA values are now allowed in the grouper using dropna=False:
pd.__version__
# '1.1.0.dev0+2004.g8d10bfb6f'
# Example from the docs
df
a b c
0 1 2.0 3
1 1 NaN 4
2 2 1.0 3
3 1 2.0 2
# without NA (the default)
df.groupby('b').sum()
a c
b
1.0 2 3
2.0 2 5
# with NA
df.groupby('b', dropna=False).sum()
a c
b
1.0 2 3
2.0 2 5
NaN 1 4
This is mentioned in the Missing Data section of the docs:
NA groups in GroupBy are automatically excluded. This behavior is consistent with R
One workaround is to use a placeholder before doing the groupby (e.g. -1):
In [11]: df.fillna(-1)
Out[11]:
a b
0 1 4
1 2 -1
2 3 6
In [12]: df.fillna(-1).groupby('b').sum()
Out[12]:
a
b
-1 2
4 1
6 3
That said, this feels pretty awful hack... perhaps there should be an option to include NaN in groupby (see this github issue - which uses the same placeholder hack).
However, as described in another answer, "from pandas 1.1 you have better control over this behavior, NA values are now allowed in the grouper using dropna=False"
Ancient topic, if someone still stumbles over this--another workaround is to convert via .astype(str) to string before grouping. That will conserve the NaN's.
df = pd.DataFrame({'a': ['1', '2', '3'], 'b': ['4', np.NaN, '6']})
df['b'] = df['b'].astype(str)
df.groupby(['b']).sum()
a
b
4 1
6 3
nan 2
I am not able to add a comment to M. Kiewisch since I do not have enough reputation points (only have 41 but need more than 50 to comment).
Anyway, just want to point out that M. Kiewisch solution does not work as is and may need more tweaking. Consider for example
>>> df = pd.DataFrame({'a': [1, 2, 3, 5], 'b': [4, np.NaN, 6, 4]})
>>> df
a b
0 1 4.0
1 2 NaN
2 3 6.0
3 5 4.0
>>> df.groupby(['b']).sum()
a
b
4.0 6
6.0 3
>>> df.astype(str).groupby(['b']).sum()
a
b
4.0 15
6.0 3
nan 2
which shows that for group b=4.0, the corresponding value is 15 instead of 6. Here it is just concatenating 1 and 5 as strings instead of adding it as numbers.
All answers provided thus far result in potentially dangerous behavior as it is quite possible you select a dummy value that is actually part of the dataset. This is increasingly likely as you create groups with many attributes. Simply put, the approach doesn't always generalize well.
A less hacky solve is to use pd.drop_duplicates() to create a unique index of value combinations each with their own ID, and then group on that id. It is more verbose but does get the job done:
def safe_groupby(df, group_cols, agg_dict):
# set name of group col to unique value
group_id = 'group_id'
while group_id in df.columns:
group_id += 'x'
# get final order of columns
agg_col_order = (group_cols + list(agg_dict.keys()))
# create unique index of grouped values
group_idx = df[group_cols].drop_duplicates()
group_idx[group_id] = np.arange(group_idx.shape[0])
# merge unique index on dataframe
df = df.merge(group_idx, on=group_cols)
# group dataframe on group id and aggregate values
df_agg = df.groupby(group_id, as_index=True)\
.agg(agg_dict)
# merge grouped value index to results of aggregation
df_agg = group_idx.set_index(group_id).join(df_agg)
# rename index
df_agg.index.name = None
# return reordered columns
return df_agg[agg_col_order]
Note that you can now simply do the following:
data_block = [np.tile([None, 'A'], 3),
np.repeat(['B', 'C'], 3),
[1] * (2 * 3)]
col_names = ['col_a', 'col_b', 'value']
test_df = pd.DataFrame(data_block, index=col_names).T
grouped_df = safe_groupby(test_df, ['col_a', 'col_b'],
OrderedDict([('value', 'sum')]))
This will return the successful result without having to worry about overwriting real data that is mistaken as a dummy value.
One small point to Andy Hayden's solution – it doesn't work (anymore?) because np.nan == np.nan yields False, so the replace function doesn't actually do anything.
What worked for me was this:
df['b'] = df['b'].apply(lambda x: x if not np.isnan(x) else -1)
(At least that's the behavior for Pandas 0.19.2. Sorry to add it as a different answer, I do not have enough reputation to comment.)
I answered this already, but some reason the answer was converted to a comment. Nevertheless, this is the most efficient solution:
Not being able to include (and propagate) NaNs in groups is quite aggravating. Citing R is not convincing, as this behavior is not consistent with a lot of other things. Anyway, the dummy hack is also pretty bad. However, the size (includes NaNs) and the count (ignores NaNs) of a group will differ if there are NaNs.
dfgrouped = df.groupby(['b']).a.agg(['sum','size','count'])
dfgrouped['sum'][dfgrouped['size']!=dfgrouped['count']] = None
When these differ, you can set the value back to None for the result of the aggregation function for that group.

How to compute mean on each column by condition

I want to compute mean for each column in a dataframe.
suppose I have a dataframe like this:
import pandas as pd
df = pd.DataFrame({'A':[1,2,3],
'B':[4,4,4],
'C':[7,8,9],
'D':[3,3,3]})
print(df)
A B C D
0 1 4 7 3
1 2 4 8 3
2 3 4 9 3
I want to get the mean as [2, 1, 8, 1]. since for column B and D only have a unique value, I hope the mean is 1, not that unique value.
How can I do that, I think I should write a function and then apply to each column by condition, but I don't know how to do that.
Hope there is anybody can help me.
That's.. a strange thing to want. :-) I'd advise against calling it a "mean", that will only confuse people.
Probably the simplest thing to do is to compute the real mean, and then just replace the unique columns with your override of 1.
In [226]: df.mean().where(df.nunique() > 1, 1)
Out[226]:
A 2.0
B 1.0
C 8.0
D 1.0
dtype: float64

Searching a particular value in a range among two columns python dataframe

I have two csv files.Depending upon the value of a cell in csv file 1 I should be able to search that value in a column of csv file 2 and get he corresponding value from other column in csv file 2.
I am sorry if this very confusing.It will probably get clear by illustration
CSV file 1
Car Mileage
A 8
B 6
C 10
CSV file 2
Score Mileage(Min) Mileage(Max)
1 1 3
2 4 6
3 7 9
4 10 12
5 13 15
And my desired output CSV file is something like this
Car Mileage Score
A 8 3
B 6 2
C 10 4
Car A is given a score of 3 depending upon its mileage 8 and then looking that mileage in csv file 2 in what range it falls and then getting corresponding score value for that range.
Any help will be appreciated
Thanks in advance
As of writing this, the current stable release is v0.21.
To read your files, use pd.read_csv -
df0 = pd.read_csv('file1.csv')
df1 = pd.read_csv('file2.csv')
df0
Car Mileage
0 A 8
1 B 6
2 C 10
df1
Score Mileage(Min) Mileage(Max)
0 1 1 3
1 2 4 6
2 3 7 9
3 4 10 12
4 5 13 15
To find the Score, use pd.IntervalIndex by calling IntervalIndex.from_tuples. This should be really fast -
v = df1.loc[:, 'Mileage(Min)':'Mileage(Max)'].apply(tuple, 1).tolist()
idx = pd.IntervalIndex.from_tuples(v, closed='both') # you can also use `from_arrays`
df0['Score'] = df1.iloc[idx.get_indexer(df0.Mileage.values), 'Score'].values
df0
Car Mileage Score
0 A 8 3
1 B 6 2
2 C 10 4
Other methods of creating an IntervalIndex are outlined here.
To write your result, use pd.DataFrame.to_csv -
df0.to_csv('file3.csv')
Here's a high level outline of what I've done here.
First, read in your CSV files
Use pd.IntervalIndex to build an interval index tree. So, searching is now logarithmic in complexity.
Use idx.get_indexer to find the index of each value in the tree
Use the index to locate the Score value in df1, and assign this back to df0. Note that I call .values, otherwise, the values will be misaligned when assigning back.
Write your result back to CSV
For more information on Intervalindex, take a look at this SO Q/A - Finding matching interval(s) in pandas Intervalindex
Note that IntervalIndex is new in v0.20, so if you have an older version, make sure you update your version with
pip install --upgrade pandas
You can use IntervalIndex, new in version 0.20.0+:
First create DataFrames by read_csv:
df1 = pd.read_csv('file1.csv')
df2 = pd.read_csv('file2.csv')
Create IntervalIndex by from_arrays:
s = pd.IntervalIndex.from_arrays(df2['Mileage(Min)'], df2['Mileage(Max)'], 'both')
print (s)
IntervalIndex([[1, 3], [4, 6], [7, 9], [10, 12], [13, 15]]
closed='both',
dtype='interval[int64]')
Select Mileage values by intervalindex and set to new column by array created by values, because else indices are not aligned and get:
TypeError: incompatible index of inserted column with frame index
df1['Score'] = df2.set_index(s).loc[df1['Mileage'], 'Score'].values
print (df1)
Car Mileage Score
0 A 8 3
1 B 6 2
2 C 10 4
And last write to file by to_csv:
df1.to_csv('file3.csv', index=False)
Setup
data = [(1,1,3), (2,4,6), (3,7,9), (4,10,12), (5,13,15)]
df = pd.DataFrame(data, columns=['Score','MMin','MMax'])
car_data = [('A', 8), ('B', 6), ('C', 10)]
car = pd.DataFrame(car_data, columns=['Car','Mileage'])
def find_score(x, df):
result = -99
for idx, row in df.iterrows():
if x >= row.MMin and x <= row.MMax:
result = row.Score
return result
car['Score'] = car.Mileage.apply(lambda x: find_score(x, df))
Which yields
In [58]: car
Out[58]:
Car Mileage Score
0 A 8 3
1 B 6 2
2 C 10 4

Apply function to pandas DataFrame that can return multiple rows

I am trying to transform DataFrame, such that some of the rows will be replicated a given number of times. For example:
df = pd.DataFrame({'class': ['A', 'B', 'C'], 'count':[1,0,2]})
class count
0 A 1
1 B 0
2 C 2
should be transformed to:
class
0 A
1 C
2 C
This is the reverse of aggregation with count function. Is there an easy way to achieve it in pandas (without using for loops or list comprehensions)?
One possibility might be to allow DataFrame.applymap function return multiple rows (akin apply method of GroupBy). However, I do not think it is possible in pandas now.
You could use groupby:
def f(group):
row = group.irow(0)
return DataFrame({'class': [row['class']] * row['count']})
df.groupby('class', group_keys=False).apply(f)
so you get
In [25]: df.groupby('class', group_keys=False).apply(f)
Out[25]:
class
0 A
0 C
1 C
You can fix the index of the result however you like
I know this is an old question, but I was having trouble getting Wes' answer to work for multiple columns in the dataframe so I made his code a bit more generic. Thought I'd share in case anyone else stumbles on this question with the same problem.
You just basically specify what column has the counts in it in and you get an expanded dataframe in return.
import pandas as pd
df = pd.DataFrame({'class 1': ['A','B','C','A'],
'class 2': [ 1, 2, 3, 1],
'count': [ 3, 3, 3, 1]})
print df,"\n"
def f(group, *args):
row = group.irow(0)
Dict = {}
row_dict = row.to_dict()
for item in row_dict: Dict[item] = [row[item]] * row[args[0]]
return pd.DataFrame(Dict)
def ExpandRows(df,WeightsColumnName):
df_expand = df.groupby(df.columns.tolist(), group_keys=False).apply(f,WeightsColumnName).reset_index(drop=True)
return df_expand
df_expanded = ExpandRows(df,'count')
print df_expanded
Returns:
class 1 class 2 count
0 A 1 3
1 B 2 3
2 C 3 3
3 A 1 1
class 1 class 2 count
0 A 1 1
1 A 1 3
2 A 1 3
3 A 1 3
4 B 2 3
5 B 2 3
6 B 2 3
7 C 3 3
8 C 3 3
9 C 3 3
With regards to speed, my base df is 10 columns by ~6k rows and when expanded is ~100,000 rows takes ~7 seconds. I'm not sure in this case if grouping is necessary or wise since it's taking all the columns to group form, but hey whatever only 7 seconds.
There is even a simpler and significantly more efficient solution.
I had to make similar modification for a table of about 3.5M rows, and the previous suggested solutions were extremely slow.
A better way is to use numpy's repeat procedure for generating a new index in which each row index is repeated multiple times according to its given count, and use iloc to select rows of the original table according to this index:
import pandas as pd
import numpy as np
df = pd.DataFrame({'class': ['A', 'B', 'C'], 'count': [1, 0, 2]})
spread_ixs = np.repeat(range(len(df)), df['count'])
spread_ixs
array([0, 2, 2])
df.iloc[spread_ixs, :].drop(columns='count').reset_index(drop=True)
class
0 A
1 C
2 C
This question is very old and the answers do not reflect pandas modern capabilities. You can use iterrows to loop over every row and then use the DataFrame constructor to create new DataFrames with the correct number of rows. Finally, use pd.concat to concatenate all the rows together.
pd.concat([pd.DataFrame(data=[row], index=range(row['count']))
for _, row in df.iterrows()], ignore_index=True)
class count
0 A 1
1 C 2
2 C 2
This has the benefit of working with any size DataFrame.

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