Python average exclude some rows - python-3.x

When i use .mean in python, it calculates the mean of the all rows.
What if I want to calculate the mean excluding the first few rows of the data ?

What I am able to understand from your question is that you have a 2d numpy array and want to calculate the mean of the rows except the first few.
You can use list slicing and reassign it to some other variable
b = a[n:]
Where new list if from nth element till last, including n.
Example:
import numpy as np
a = np.array([[1, 2], [3, 4]])
b = a[1:]
print(a.mean(), b.mean())
Output:
2.5 3.5

Related

How can I apply a function to multiple columns of grouped rows and make a column of the output?

I have a large excel file containing stock data loaded from and sorted from an API. Below is a sample of dummy data that is applicable towards solving this problem:
I would like to create a function that Scores Stocks grouped by their Industry. For this sample I would like to score a stock a 5 if its ['Growth'] value is less than its groups mean Growth (quantile, percentile) in Growth and a 10 if above the groups mean growth. The scores of all values in a column should be returned in a list
Current Unfinished Code:
import numpy as np
import pandas as pd
data = pd.DataFrame(pd.read_excel('dummydata.xlsx')
Desired input:
data['Growth'].apply(score) # Scores stock
Desired Output:
[5, 10, 10, 5, 10, 5]
If I can create a function for this sample then I will be able to make similar ones for different columns with slightly different conditions and aggregates (like percentile or quantile) that affect the score. I'd say the main problem here is accessing these grouped values and comparing them.
I don't think it's possible to convert from a Series to a list in the apply call. I may be wrong on that but if the desired output was changed slightly to
data['Growth'].apply(score).tolist()
then you can use a lambda function to do this.
score = lambda x: 5 if x < growth.mean() else 10
data['Growth'].apply(score).tolist() # outputs [5, 10, 10, 5, 10, 5]

How to use dataframe column in for loop

I am trying to implement a formula to create a new column in Dataframe using existing column but that column is a summation from 0 to a number present in some other column.
I was trying something like this:
dataset['B']=sum([1/i for i in range(dataset['A'])])
I know something like this would work
dataset['B']=sum([1/i for i in range(10)])
but I want to make this 10 dynamic based on some different column.
I keep on getting this error.
TypeError: 'Series' object cannot be interpreted as an integer
First of all, I should admit that I could not understand you question completely. However, what I understood that you want to iterate over the rows of a DataFrame and make a new column by doing some operation/s on that value.
Is that is so, then I would recommend you following link
Regarding TypeError: 'Series' object cannot be interpreted as an integer:
The init signature range() takes integers as input. i.e [i for i in range(10)] should give you [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. However, if one of the value from your dataset['A'] is float, or not integer , this might result in the error you are having. Moreover, if you notice, the first value is a zero, as a result, 1/i should result in a different error. As a result, you might have to rewrite the code as [1/i for i in range (1 , row_value_of_dataset['A'])]
It will be highly appreciate if you could make an example of what you DataFrame might look like and what is your desired output. Then perhaps it is easier to post a solution.
BTW forgot to post what I understood from your question:
#assume the data:
>>>import pandas as pd
>>>data = pd.DataFrame({'A': (1, 2, 3, 4)})
#the data
>>>data
A
0 1
1 2
2 3
3 4
#doing operation on each of the rows
>>>data['B']=data.apply(lambda row: sum([1/i for i in range(1, row.A)] ), axis=1)
# Column B is the newly added data
>>>data
A B
0 1 0.000000
1 2 1.000000
2 3 1.500000
3 4 1.833333
Perhaps explicitly use cumsum, or even apply?
Anyway trying to move an array/list item directly into a dataframe and seem to view this as a dictionary. Try something like this, I've not tested it,
array_x = [x, 1/x for x in dataset.values.tolist()] # or `dataset.A.tolist()`
df = pd.DataFrame(data=(np.asarray(array_x)))
df.columns = [A, B]
Here the idea is to break the Series apart into a list, and input the list into a dataframe. This can be explicitly done without needing to go Series->list->dataframe and is not very efficient.

P-value normal test for multiple rows

I got the following simple code to calculate normality over an array:
import pandas as pd
df = pd.read_excel("directory\file.xlsx")
import numpy as np
x=df.iloc[:,1:].values.flatten()
import scipy.stats as stats
from scipy.stats import normaltest
stats.normaltest(x,axis=None)
This gives me nicely a p-value and a statistic.
The only thing I want right now is to:
Add 2 columns in the file with this p value and statistic and if i have multiple rows, do it for all the rows (calculate p value & statistic for each row and add 2 columns with these values in it).
Can someone help?
If you want to calculate row-wise normaltest, you should not flatten your data in x and use axis=1 such as
df = pd.DataFrame(np.random.random(105).reshape(5,21)) # to generate data
# calculate normaltest row-wise without the first column like you
df['stat'] ,df['p'] = stats.normaltest(df.iloc[:,1:],axis=1)
Then df contains two columns 'stat' and 'p' with the values your are looking for IIUC.
Note: to be able to perform normaltest, you need at least 8 values (according to what I experienced) so you need at least 8 columns in df.iloc[:,1:] otherwise it will raise an error. And even, it would be better to have more than 20 values in each row.

Python Pandas: Get index of rows which column matches certain value (max) [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.

Pandas sort not maintaining sort

What is the right way to multiply two sorted pandas Series?
When I run the following
import pandas as pd
x = pd.Series([1,3,2])
x.sort()
print(x)
w = [1]*3
print(w*x)
I get what I would expect - [1,2,3]
However, when I change it to a Series:
w = pd.Series(w)
print(w*x)
It appears to multiply based on the index of the two series, so it returns [1,3,2]
Your results are essentially the same, just sorted differently.
>>> w*x
0 1
2 2
1 3
>>> pd.Series(w)*x
0 1
1 3
2 2
>>> (w*x).sort_index()
0 1
1 3
2 2
The rule is basically this: Anytime you multiply a dataframe or series by a dataframe or series, it will be done by index. That's what makes it pandas and not numpy. As a result, any pre-sorting is necessarily ignored.
But if you multiply a dataframe or series by a list or numpy array of a conforming shape/size, then the list or array will be treated as having the exact same index as the dataframe or series. The pre-sorting of the series or dataframe can be preserved in this case because there can not be any conflict with the list or array (which don't have an index at all).
Both of these types of behavior can be very desirable depending on what you are trying do. That's why you will often see answers here that do something like df1 * df2.values when the second type of behavior is desired.
In this example, it doesn't really matter because your list is [1,1,1] and gives the same answer either way, but if it was [1,2,3] you would get different answers, not just differently sorted answers.

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