It seems simple but I can’t seem to find an efficient way to solve this in Python 3: Is there is a loop I can use in my dataframe that takes every column after the current column (starting with the 1st column), and subtracts it from the current column, so that I can add that resulting column to a new dataframe?
This is what my data looks like:
This is what I have so far, but when running run_analysis my "result" equation is bringing up an error, and I do not know how to store the results in a new dataframe. I'm a beginner at all of this so any help would be much appreciated.
storage = [] #container that will store the results of the subtracted columns
def subtract (a,b): #function to call to do the column-wise subtractions
return a-b
def run_analysis (frame, store):
for first_col_index in range(len(frame)): #finding the first column to use
temp=[] #temporary place to store the column-wise values from the analysis
for sec_col_index in range(len(frame)): #finding the second column to subtract from the first column
if (sec_col_index <= first_col_index): #if the column is below the current column or is equal to
#the current column, then skip to next column
continue
else:
result = [r for r in map(subtract, frame[sec_col_index], frame[first_col_index])]
#if column above our current column, the subtract values in the column and keep the result in temp
temp.append(result)
store.append(temp) #save the complete analysis in the store
Something like this?
#dummy ddataframe
df = pd.DataFrame({'a':list(range(10)), 'b':list(range(10,20)), 'c':list(range(10))})
print(df)
output:
a b c
0 0 10 0
1 1 11 1
2 2 12 2
3 3 13 3
4 4 14 4
5 5 15 5
6 6 16 6
7 7 17 7
8 8 18 8
9 9 19 9
Now iterate over pairs of columns and subtract them while assigning another column to the dataframe
for c1, c2 in zip(df.columns[:-1], df.columns[1:]):
df[f'{c2}-{c1}'] = df[c2]-df[c1]
print(df)
output:
a b c b-a c-b
0 0 10 0 10 -10
1 1 11 1 10 -10
2 2 12 2 10 -10
3 3 13 3 10 -10
4 4 14 4 10 -10
5 5 15 5 10 -10
6 6 16 6 10 -10
7 7 17 7 10 -10
8 8 18 8 10 -10
9 9 19 9 10 -10
Related
I have a pandas dataframe and I would like to add a column and set certain values based on certain conditions.
Initialization
See my following example:
import numpy as np
import pandas as pd
df=pd.DataFrame(np.reshape(np.arange(20),(5,4)).tolist(),columns=list('ABCD'))
df['E']=''
df.loc[0,'E']=1
df
Until here, everything is as wished. Output:
A B C D E
0 0 1 2 3 1
1 4 5 6 7
2 8 9 10 11
3 12 13 14 15
4 16 17 18 19
Problem
But now, I would like to set the value of the 'E' column in the first row for which the 'B' value is larger than 8 to 3 (the same as df.loc[2,'E']=1 but using that condition. I don't know which line it is a priori).
Solution attempts
I have tried different things:
df.loc[df['B']>8,'E'].iloc[0]=3
df
This doesn't change any value:
A B C D E
0 0 1 2 3 1
1 4 5 6 7
2 8 9 10 11
3 12 13 14 15
4 16 17 18 19
The same holds for
df.loc[df['B']>8]['E'].iloc[0]=3
Other things I've tried result in an error. df.loc[df['B']>8,'E']=3works well but I only want to assign the value in the first line meeting the condition, not in all lines.
So how do I achieve this?
Use .idxmax(). Since True is the max of True/False it will return the index of the first True value for a boolean index.
df.loc[(df['B']>8).idxmax(),'E']=3
I'm having problems with pd.rolling() method that returns several outputs even though the function returns a single value.
My objective is to:
Calculate the absolute percentage difference between two DataFrames with 3 columns in each df.
Sum all values
I can do this using pd.iterrows(). But working with larger datasets makes this method ineffective.
This is the test data im working with:
#import libraries
import pandas as pd
import numpy as np
#create two dataframes
values = {'column1': [7,2,3,1,3,2,5,3,2,4,6,8,1,3,7,3,7,2,6,3,8],
'column2': [1,5,2,4,1,5,5,3,1,5,3,5,8,1,6,4,2,3,9,1,4],
"column3" : [3,6,3,9,7,1,2,3,7,5,4,1,4,2,9,6,5,1,4,1,3]
}
df1 = pd.DataFrame(values)
df2 = pd.DataFrame([[2,3,4],[3,4,1],[3,6,1]])
print(df1)
print(df2)
column1 column2 column3
0 7 1 3
1 2 5 6
2 3 2 3
3 1 4 9
4 3 1 7
5 2 5 1
6 5 5 2
7 3 3 3
8 2 1 7
9 4 5 5
10 6 3 4
11 8 5 1
12 1 8 4
13 3 1 2
14 7 6 9
15 3 4 6
16 7 2 5
17 2 3 1
18 6 9 4
19 3 1 1
20 8 4 3
0 1 2
0 2 3 4
1 3 4 1
2 3 6 1
This method produces the output I want by using pd.iterrows()
RunningSum = []
for index, rows in df1.iterrows():
if index > 3:
Div = abs((((df2 / df1.iloc[index-3+1:index+1].reset_index(drop="True").values)-1)*100))
Average = Div.sum(axis=0)
SumOfAverages = np.sum(Average)
RunningSum.append(SumOfAverages)
#printing my desired output values
print(RunningSum)
[991.2698412698413,
636.2698412698412,
456.19047619047626,
616.6666666666667,
935.7142857142858,
627.3809523809524,
592.8571428571429,
350.8333333333333,
449.1666666666667,
1290.0,
658.531746031746,
646.031746031746,
597.4603174603175,
478.80952380952385,
383.0952380952381,
980.5555555555555,
612.5]
Finally, below is my attemt to use pd.rolling() so that I dont need to loop through each row.
def SumOfAverageFunction(vals):
Div = abs((((df2.values / vals.reset_index(drop="True").values)-1)*100))
Average = Div.sum()
SumOfAverages = np.sum(Average)
return SumOfAverages
RunningSums = df1.rolling(window=3,axis=0).apply(SumOfAverageFunction)
Here is my problem because printing RunningSums from above outputs several values and is not close to the results I'm getting using iterrows method. How do I solve this?
print(RunningSums)
column1 column2 column3
0 NaN NaN NaN
1 NaN NaN NaN
2 702.380952 780.000000 283.333333
3 533.333333 640.000000 533.333333
4 1200.000000 475.000000 403.174603
5 833.333333 1280.000000 625.396825
6 563.333333 760.000000 1385.714286
7 346.666667 386.666667 1016.666667
8 473.333333 573.333333 447.619048
9 533.333333 1213.333333 327.619048
10 375.000000 746.666667 415.714286
11 408.333333 453.333333 515.000000
12 604.166667 338.333333 1250.000000
13 1366.666667 577.500000 775.000000
14 847.619048 1400.000000 683.333333
15 314.285714 733.333333 455.555556
16 533.333333 441.666667 474.444444
17 347.619048 616.666667 546.666667
18 735.714286 466.666667 1290.000000
19 350.000000 488.888889 875.000000
20 525.000000 1361.111111 1266.666667
It's just the way rolling behaves, it's going to window around all of the columns and I don't know that there is a way around it. One solution is to apply rolling to a single column, and use the indexes from those windows to slice the dataframe inside your function. Still expensive, but probably not as bad as what you're doing.
Also the output of your first method looks wrong. You're actually starting your calculations a few rows too late.
import numpy as np
def SumOfAverageFunction(vals):
return (abs(np.divide(df2.values, df1.loc[vals.index].values)-1)*100).sum()
vals = df1.column1.rolling(3)
vals.apply(SumOfAverageFunction, raw=False)
This is my desired output:
I am trying to calculate the column df[Value] and df[Value_Compensed]. However, to do that, I need to consider the previous value of the row df[Value_Compensed]. In terms of my table:
The first row all the values are 0
The following rows: df[Remained] = previous df[Value_compensed]. Then df[Value] = df[Initial_value] + df[Remained]. Then df[Value_Compensed] = df[Value] - df[Compensation]
...And So on...
I am struggling to pass the value of Value_Compensed from one row to the next, I tried with the function shift() but as you can see in the following image the values in df[Value_Compensed] are not correct due to it is not a static value and also it also changes after each row it did not work. Any Idea??
Thanks.
Manuel.
You can use apply to create your customised operations. I've made a dummy dataset as you didn't provide the initial dataframe.
from itertools import zip_longest
# dummy data
df = pd.DataFrame(np.random.randint(1, 10, (8, 5)),
columns=['compensation', 'initial_value',
'remained', 'value', 'value_compensed'],)
df.loc[0] = 0,0,0,0,0
>>> print(df)
compensation initial_value remained value value_compensed
0 0 0 0 0 0
1 2 9 1 9 7
2 1 4 9 8 3
3 3 4 5 7 6
4 3 2 5 5 6
5 9 1 5 2 4
6 4 5 9 8 2
7 1 6 9 6 8
Use apply (axis=1) to do row-wise iteration, where you use the initial dataframe as an argument, from which you can then get the previous row x.name-1 and do your calculations. Not sure if I fully understood the intended result, but you can adjust the individual calculations of the different columns in the function.
def f(x, data):
if x.name == 0:
return [0,]*data.shape[1]
else:
x_remained = data.loc[x.name-1]['value_compensed']
x_value = data.loc[x.name-1]['initial_value'] + x_remained
x_compensed = x_value - x['compensation']
return [x['compensation'], x['initial_value'], x_remained, \
x_value, x_compensed]
adj = df.apply(f, args=(df,), axis=1)
adj = pd.DataFrame.from_records(zip_longest(*adj.values), index=df.columns).T
>>> print(adj)
compensation initial_value remained value value_compensed
0 0 0 0 0 0
1 5 9 0 0 -5
2 5 7 4 13 8
3 7 9 1 8 1
4 6 6 5 14 8
5 4 9 6 12 8
6 2 4 2 11 9
7 9 2 6 10 1
I have a large time series df (2.5mil rows) that contain 0 values in a given row, some of which are legitimate. However if there are repeated continuous occurrences of zero values I would like to remove them from my df.
Example:
Col. A contains [1,2,3,0,4,5,0,0,0,1,2,3,0,8,8,0,0,0,0,9] I would like to remove the [0,0,0] and [0,0,0,0] from the middle and leave the remaining 0 to make a new df [1,2,3,0,4,5,1,2,3,0,8,8,9].
The length of zero values before deletion being a parameter that has to be set - in this case > 2.
Is there a clever way to do this in pandas?
It looks like you want to remove the row if it is 0 and either previous or next row in same column is 0. You can use shift to look for previous and next value and compare with current value as below:
result_df = df[~(((df.ColA.shift(-1) == 0) & (df.ColA == 0)) | ((df.ColA.shift(1) == 0) & (df.ColA == 0)))]
print(result_df)
Result:
ColA
0 1
1 2
2 3
3 0
4 4
5 5
9 1
10 2
11 3
12 0
13 8
14 8
19 9
Update for more than 2 consecutive
Following example in link, adding new column to track consecutive occurrence and later checking it to filter:
# https://stackoverflow.com/a/37934721/5916727
df['consecutive'] = df.ColA.groupby((df.ColA != df.ColA.shift()).cumsum()).transform('size')
df[~((df.consecutive>10) & (df.ColA==0))]
We need build a new para meter here, then using drop_duplicates
df['New']=df.A.eq(0).astype(int).diff().ne(0).cumsum()
s=pd.concat([df.loc[df.A.ne(0),:],df.loc[df.A.eq(0),:].drop_duplicates(keep=False)]).sort_index()
s
Out[190]:
A New
0 1 1
1 2 1
2 3 1
3 0 2
4 4 3
5 5 3
9 1 5
10 2 5
11 3 5
12 0 6
13 8 7
14 8 7
19 9 9
Explanation :
#df.A.eq(0) to find the value equal to 0
#diff().ne(0).cumsum() if they are not equal to 0 then we would count them in same group .
I am new to using pandas but want to learn it better. I am currently facing a problem. I have a DataFrame looking like this:
0 1 2
0 chr2L 1 4
1 chr2L 9 12
2 chr2L 17 20
3 chr2L 23 23
4 chr2L 26 27
5 chr2L 30 40
6 chr2L 45 47
7 chr2L 52 53
8 chr2L 56 56
9 chr2L 61 62
10 chr2L 66 80
I want to get something like this:
0 1 2 3
0 chr2L 0 1 0
1 chr2L 1 2 1
2 chr2L 2 3 1
3 chr2L 3 4 1
4 chr2L 4 5 0
5 chr2L 5 6 0
6 chr2L 6 7 0
7 chr2L 7 8 0
8 chr2L 8 9 0
9 chr2L 9 10 1
10 chr2L 10 11 1
11 chr2L 11 12 1
12 chr2L 12 13 0
And so on...
So, fill in the missing intervals with zeros, and save the present intervals as ones (if there is an easy way to save "boundary" positions (the borders of the intervals in the initial data) as 0.5 at the same time it might also be helpful) while splitting all data into 1-length intervals.
In the data there are multiple string values in the column 0, and this should be done for each of them separately. They require different length of the final data (the last value that should get a 0 or a 1 is different). Would appreciate your help with dealing with this in pandas.
This works for most of your first paragraph and some of the second. Left as an exercise: finish inserting insideness=0 rows (see end):
import pandas as pd
# dummied-up version of your data, but with column headers for readability:
df = pd.DataFrame({'n':['a']*4 + ['b']*2, 'a':[1,6,8,5,1,5],'b':[4,7,10,5,3,7]})
# splitting up a range, translated into df row terms:
def onebyone(dfrow):
a = dfrow[1].a; b = dfrow[1].b; n = dfrow[1].n
count = b - a
if count >= 2:
interior = [0.5]+[1]*(count-2)+[0.5]
elif count == 1:
interior = [0.5]
elif count == 0:
interior = []
return {'n':[n]*count, 'a':range(a, a + count),
'b':range(a + 1, a + count + 1),
'insideness':interior}
Edited to use pd.concat(), new in pandas 0.15, to combine the intermediate results:
# Into a new dataframe:
intermediate = []
for label in set(df.n):
for row in df[df.n == label].iterrows():
intermediate.append(pd.DataFrame(onebyone(row)))
df_onebyone = pd.concat(intermediate)
df_onebyone.index = range(len(df_onebyone))
And finally a sketch of identifying the missing rows, which you can edit to match the above for-loop in adding rows to a final dataframe:
# for times in the overall range describing 'a'
for i in range(int(newd[newd.n=='a'].a.min()),int(newd[newd.n=='a'].a.max())):
# if a time isn't in an existing 0.5-1-0.5 range:
if i not in newd[newd.n=='a'].a.values:
# these are the values to fill in a 0-row
print '%d, %d, 0'%(i, i+1)
Or, if you know the a column will be sorted for each n, you could keep track of the last end-value handled by onebyone() and insert some extra rows to catch up to the next start value you're going to pass to onebyone().