How to iterate and calculate over a pandas multi-index dataframe - python

I have a pandas multi-index dataframe:
>>> df
0 1
first second
A one 0.991026 0.734800
two 0.582370 0.720825
B one 0.795826 -1.155040
two 0.013736 -0.591926
C one -0.538078 0.291372
two 1.605806 1.103283
D one -0.617655 -1.438617
two 1.495949 -0.936198
I'm trying to find an efficient way to divide each number in column 0 by the maximum number in column I that shares the same group under index "first", and make this into a third column. Is there a simply efficient method for doing something like this that doesn't require multiple for loops?

Use Series.div with max for maximal values per first level:
print (df[1].max(level=0))
first
A 0.734800
B -0.591926
C 1.103283
D -0.936198
Name: 1, dtype: float64
df['new'] = df[0].div(df[1].max(level=0))
print (df)
0 1 new
first second
A one 0.991026 0.734800 1.348702
two 0.582370 0.720825 0.792556
B one 0.795826 -1.155040 -1.344469
two 0.013736 -0.591926 -0.023206
C one -0.538078 0.291372 -0.487706
two 1.605806 1.103283 1.455480
D one -0.617655 -1.438617 0.659748
two 1.495949 -0.936198 -1.597898

Related

Dataframe- Remove similar rows related based on two columns

I have following dataset:
this dataset print correlation of two columns at left
if you look at the row number 3 and 42, you will find they are same. only column position is different. that does not affect correlation. I want to remove column 42. But this dataset has many these row of similar values. I need a general algorithm to remove these similar value and have only unique.
As the correlation_value seems to be the same, the operation should be commutative, so whatever the value, you just have to focus on two first columns. Sort the tuple and remove duplicates
# You can probably replace 'sorted' by 'set'
key = df[['source_column', 'destination_column']] \
.apply(lambda x: tuple(sorted(x)), axis='columns')
out = df.loc[~key.duplicated()]
>>> out
source_column destination_column correlation_Value
0 A B 1
2 C E 2
3 D F 4
You could try a self join. Without a code example, it's hard to answer, but something like this maybe:
df.merge(df, left_on="source_column", right_on="destination_column")
You can follow that up with a call to drop_duplicates.

Returning date that corresponds with maximum value in pandas dataframe [duplicate]

How can I find the row for which the value of a specific column is maximal?
df.max() will give me the maximal value for each column, I don't know how to get the corresponding row.
Use the pandas idxmax function. It's straightforward:
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1
Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.
idxmax() returns indices labels, not integers.
Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').
if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).
HISTORICAL NOTES:
idxmax() used to be called argmax() prior to 0.11
argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
argmax function returned the integer position within the index of the row location of the maximum element.
pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.
For example, consider this toy DataFrame with a duplicate row label:
In [19]: dfrm
Out[19]:
A B C
a 0.143693 0.653810 0.586007
b 0.623582 0.312903 0.919076
c 0.165438 0.889809 0.000967
d 0.308245 0.787776 0.571195
e 0.870068 0.935626 0.606911
f 0.037602 0.855193 0.728495
g 0.605366 0.338105 0.696460
h 0.000000 0.090814 0.963927
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
In [20]: dfrm['A'].idxmax()
Out[20]: 'i'
In [21]: dfrm.iloc[dfrm['A'].idxmax()] # .ix instead of .iloc in older versions of pandas
Out[21]:
A B C
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).
This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.
So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.
You might also try idxmax:
In [5]: df = pandas.DataFrame(np.random.randn(10,3),columns=['A','B','C'])
In [6]: df
Out[6]:
A B C
0 2.001289 0.482561 1.579985
1 -0.991646 -0.387835 1.320236
2 0.143826 -1.096889 1.486508
3 -0.193056 -0.499020 1.536540
4 -2.083647 -3.074591 0.175772
5 -0.186138 -1.949731 0.287432
6 -0.480790 -1.771560 -0.930234
7 0.227383 -0.278253 2.102004
8 -0.002592 1.434192 -1.624915
9 0.404911 -2.167599 -0.452900
In [7]: df.idxmax()
Out[7]:
A 0
B 8
C 7
e.g.
In [8]: df.loc[df['A'].idxmax()]
Out[8]:
A 2.001289
B 0.482561
C 1.579985
Both above answers would only return one index if there are multiple rows that take the maximum value. If you want all the rows, there does not seem to have a function.
But it is not hard to do. Below is an example for Series; the same can be done for DataFrame:
In [1]: from pandas import Series, DataFrame
In [2]: s=Series([2,4,4,3],index=['a','b','c','d'])
In [3]: s.idxmax()
Out[3]: 'b'
In [4]: s[s==s.max()]
Out[4]:
b 4
c 4
dtype: int64
df.iloc[df['columnX'].argmax()]
argmax() would provide the index corresponding to the max value for the columnX. iloc can be used to get the row of the DataFrame df for this index.
A more compact and readable solution using query() is like this:
import pandas as pd
df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
print(df)
# find row with maximum A
df.query('A == A.max()')
It also returns a DataFrame instead of Series, which would be handy for some use cases.
Very simple: we have df as below and we want to print a row with max value in C:
A B C
x 1 4
y 2 10
z 5 9
In:
df.loc[df['C'] == df['C'].max()] # condition check
Out:
A B C
y 2 10
If you want the entire row instead of just the id, you can use df.nlargest and pass in how many 'top' rows you want and you can also pass in for which column/columns you want it for.
df.nlargest(2,['A'])
will give you the rows corresponding to the top 2 values of A.
use df.nsmallest for min values.
The direct ".argmax()" solution does not work for me.
The previous example provided by #ely
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].argmax()
3
>>> df['B'].argmax()
4
>>> df['C'].argmax()
1
returns the following message :
FutureWarning: 'argmax' is deprecated, use 'idxmax' instead. The behavior of 'argmax'
will be corrected to return the positional maximum in the future.
Use 'series.values.argmax' to get the position of the maximum now.
So that my solution is :
df['A'].values.argmax()
mx.iloc[0].idxmax()
This one line of code will give you how to find the maximum value from a row in dataframe, here mx is the dataframe and iloc[0] indicates the 0th index.
Considering this dataframe
[In]: df = pd.DataFrame(np.random.randn(4,3),columns=['A','B','C'])
[Out]:
A B C
0 -0.253233 0.226313 1.223688
1 0.472606 1.017674 1.520032
2 1.454875 1.066637 0.381890
3 -0.054181 0.234305 -0.557915
Assuming one want to know the rows where column "C" is max, the following will do the work
[In]: df[df['C']==df['C'].max()])
[Out]:
A B C
1 0.472606 1.017674 1.520032
The idmax of the DataFrame returns the label index of the row with the maximum value and the behavior of argmax depends on version of pandas (right now it returns a warning). If you want to use the positional index, you can do the following:
max_row = df['A'].values.argmax()
or
import numpy as np
max_row = np.argmax(df['A'].values)
Note that if you use np.argmax(df['A']) behaves the same as df['A'].argmax().
Use:
data.iloc[data['A'].idxmax()]
data['A'].idxmax() -finds max value location in terms of row
data.iloc() - returns the row
If there are ties in the maximum values, then idxmax returns the index of only the first max value. For example, in the following DataFrame:
A B C
0 1 0 1
1 0 0 1
2 0 0 0
3 0 1 1
4 1 0 0
idxmax returns
A 0
B 3
C 0
dtype: int64
Now, if we want all indices corresponding to max values, then we could use max + eq to create a boolean DataFrame, then use it on df.index to filter out indexes:
out = df.eq(df.max()).apply(lambda x: df.index[x].tolist())
Output:
A [0, 4]
B [3]
C [0, 1, 3]
dtype: object
what worked for me is:
df[df['colX'] == df['colX'].max()
You then get the row in your df with the maximum value of colX.
Then if you just want the index you can add .index at the end of the query.

Python: How to pass row and next row DataFrame.apply() method?

I have DataFrame with thousands rows. Its structure is as below
A B C D
0 q 20 'f'
1 q 14 'd'
2 o 20 'a'
I want to compare the A column of current row and next row. If those values are equal I want to add the value of B column which has lower the value to D column of compared row which has greater value. Then I want to remove the moved column value of column B. It's like a swap process.
A B C D
0 q 20 'f' 14
1 o 20 'a'
I have thousands rows and iloc, loc, at methods work slow. At least I want to use DataFrame apply method. I tried some code samples but they didn't work.
I want to do something as below:
DataFrame.apply(lambda row: self.compare(row, next(row)), axis=1))
I have a compare method but I couldn't pass next row to the compare method. How can I pass it to the method? Also I am open to hear faster pandas solutions.
Best not to do that with apply as it will be slow; you can look at using shift, e.g.
df['A_shift'] = df['A'].shift(1)
df['Is_Same'] = 0
df.loc[df.A_shift == df.A, 'Is_Same'] = 1
Gets a bit more complicated if you're doing the shift within groups, but still possible.

How to get the mean of a subset of rows after using groupby?

I want to get the average of a particular subset of rows in one particular column in my dataframe.
I can use
df['C'].iloc[2:9].mean()
to get the mean of just the particular rows I want from my original Dataframe but my problem is that I want to perform this operation after using the groupby operation.
I am building on
df.groupby(["A", "B"])['C'].mean()
whereby there are 11 values returned in 'C' once I group by columns A and B and I get the average of those 11 values. I actually only want to get the average of the 3rd through 9th values though so ideally what I would want to do is
df.groupby(["A", "B"])['C'].iloc[2:9].mean()
This would return those 11 values from column C for every group of A,B and then would find the mean of the 3rd through 9th values but I know I can't do this. The error suggests using the apply method but I can't seem to figure it out.
Any help would be appreciated.
You can use agg function after the groupby and then subset within each group and take the mean:
df = pd.DataFrame({'A': ['a']*22, 'B': ['b1']*11 + ['b2']*11, 'C': list(range(11))*2})
# A dummy data frame to demonstrate
df.groupby(['A', 'B'])['C'].agg(lambda g: g.iloc[2:9].mean())
# A B
# a b1 5
# b2 5
# Name: C, dtype: int64
Try this variant:
for key, grp in df.groupby(["A", "B"]):
print grp['C'].iloc[2:9].mean()

Get integer row index of MultiIndex Series

I have a pandas Series with a MultiIndex, and I want to get the integer row numbers that belong to one level of the MultiIndex.
For example, if I have sample data s
s = pandas.Series([10, 23, 2, 19],
index=pandas.MultiIndex.from_product([['a', 'b'], ['c', 'd']]))
which looks like this:
a c 10
d 23
b c 2
d 19
I want to get the row numbers that correspond to the level b. So here, I'd get [2, 3] as the output, because the last two rows are under b. Also, I really only need the first row that belongs under b.
I wanted to get the numbers so that I can compare across Series. Say I have five Series objects with a b level. These are time-series data, and b corresponds to a condition that was present during some of the observations (and c is a sub-condition, etc). I want to see which Series had the conditions present at the same time.
Edit: To clarify, I don't need to compare the values themselves, just the indices. For example, in R if I had this dataframe:
d = data.frame(col_1 = c('a','a','b','b'), col_2 = c('c','d','c','d'), col_3 = runif(4))
Then the command which(d$col_1 == 'b') would produce the results I want.
If the index that you want to index by is the outermost one you can use loc
df.loc['b']
To get the first row I find the head method the easiest
df.loc['b'].head(1)
The idiomatic way to do the second part of your question is as follows. Say your series are named series1, series2 and series3.
big_series = pd.concat([series1, series2, series3])
big_series.loc['b']

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