Follow-up rolling_apply deprecated - python

Following up on this answer: Is there a way to do a weight-average rolling sum over a grouping?
rsum = pd.rolling_apply(g.values,p,lambda x: np.nansum(w*x),min_periods=p)
rolling_apply is deprecated now. How would you change this to work under current functionality.

As of 0.18+, use Series.rolling.apply.
w = np.array([0.1,0.1,0.2,0.6])
df.groupby('ID').VALUE.apply(
lambda x: x.rolling(window=4).apply(lambda x: np.dot(x, w), raw=False))
0 NaN
1 NaN
2 NaN
3 146.0
4 166.0
5 NaN
6 NaN
7 NaN
8 2.5
9 NaN
10 NaN
11 NaN
12 35.5
13 21.4
14 NaN
15 NaN
16 NaN
17 8.3
18 9.8
19 NaN
Name: VALUE, dtype: float64
The raw argument is new to 0.23 (set it to specify passing Series v/s arrays), so remove it if you're having trouble on older versions.

Related

Replace Unnamed values in date column with true values

I'm working on this raw data frame that needs some cleaning. So far, I have transformed this xlsx file
into this pandas dataframe:
print(df.head(16))
date technician alkalinity colour uv ph turbidity \
0 2020-02-01 00:00:00 Catherine 24.5 33 0.15 7.24 1.53
1 Unnamed: 2 NaN NaN NaN NaN NaN 2.31
2 Unnamed: 3 NaN NaN NaN NaN NaN 2.08
3 Unnamed: 4 NaN NaN NaN NaN NaN 2.2
4 Unnamed: 5 Michel 24 35 0.152 7.22 1.59
5 Unnamed: 6 NaN NaN NaN NaN NaN 1.66
6 Unnamed: 7 NaN NaN NaN NaN NaN 1.71
7 Unnamed: 8 NaN NaN NaN NaN NaN 1.53
8 2020-02-02 00:00:00 Catherine 24 NaN 0.145 7.21 1.44
9 Unnamed: 10 NaN NaN NaN NaN NaN 1.97
10 Unnamed: 11 NaN NaN NaN NaN NaN 1.91
11 Unnamed: 12 NaN NaN 33.0 NaN NaN 2.07
12 Unnamed: 13 Michel 24 34 0.15 7.24 1.76
13 Unnamed: 14 NaN NaN NaN NaN NaN 1.84
14 Unnamed: 15 NaN NaN NaN NaN NaN 1.72
15 Unnamed: 16 NaN NaN NaN NaN NaN 1.85
temperature
0 3
1 NaN
2 NaN
3 NaN
4 3
5 NaN
6 NaN
7 NaN
8 3
9 NaN
10 NaN
11 NaN
12 3
13 NaN
14 NaN
15 NaN
From here, I want to combine the rows so that I only have one row for each date. The values for each row will be the mean in the respective columns. ie.
print(new_df.head(2))
date time alkalinity colour uv ph turbidity temperature
0 2020-02-01 00:00:00 24.25 34 0.151 7.23 1.83 3
1 2020-02-02 00:00:00 24 33.5 0.148 7.23 1.82 3
How can I accomplish this when I have Unnamed values in my date column? Thanks!
Try setting the values to NaN and then use ffill:
df.loc[df.date.str.contains('Unnamed', na=False), 'date'] = np.nan
df.date = df.date.ffill()
If I understand, correctly you want to drop rows that contain 'Unnamed' in the date column, right?
Please look here:
https://stackoverflow.com/a/27360130/12790501
The solution would be something like this:
df = df.drop(df['Unnamed' in df.date].index)
Edit:
No, I would like to replace those Unnamed values with the date so I
could then use the groupby('date') function to return the mean values
for the columns
so in the case you should just iterate over the whole table
last_date = ''
for i in df.index:
if 'Unnamed' not in df.at[i, 'date']:
last_date = df.at[i, 'date']
else:
df.at[i, 'date'] = last_date
If the 'date' column is of type object i.e. string
then just write a logic to loop over the number as seen in image provided it follows a certain pattern-
for _ in range(2,9):
df.loc[(df['date'] == 'Unnamed: '+str(_), 'date'] = your_value

Getting corresponding values in a groupby

I have a dataset similar to this
Serial A B
1 12
1 31
1
1 12
1 31 203
1 10
1 2
2 32 100
2 32 242
2 3
3 2
3 23 100
3
3 23
I group the dataframe based on Serial and find the maximum value of each A column by df['A_MAX'] = df.groupby('Serial')['A'].transform('max').values and retain the first value by df['A_MAX'] = df['A_MAX'].mask(df['Serial'].duplicated(), '')
Serial A B A_MAX B_corresponding
1 12 31 203
1 31
1
1 12
1 31 203
1 10
1 2
2 32 100 32 100
2 32 242
2 3
3 2 23 100
3 23 100
3
3 23
Now for the B_corresponding column, I would like to get the corresponding B values of the A_MAX. I thought of locating the A_MAX values in A but there are similar max A values per group. Additional condition, for example in Serial 2 I would also prefer to get the smallest B values between the 32
Idea is use DataFrame.sort_values for maximal values per groups, then remove missing values by DataFrame.dropna and get first rows by Serial by DataFrame.drop_duplicates. Create Series by DataFrame.set_index and last use Series.map:
df['A_MAX'] = df.groupby('Serial')['A'].transform('max')
df['A_MAX'] = df['A_MAX'].mask(df['Serial'].duplicated())
s = (df.sort_values(['Serial','A'], ascending=[True, False])
.dropna(subset=['B'])
.drop_duplicates('Serial')
.set_index('Serial')['B'])
df['B_corresponding'] = df['Serial'].map(s).mask(df['Serial'].duplicated())
print (df)
Serial A B A_MAX B_corresponding
0 1 12.0 NaN 31.0 203.0
1 1 31.0 NaN NaN NaN
2 1 NaN NaN NaN NaN
3 1 12.0 NaN NaN NaN
4 1 31.0 203.0 NaN NaN
5 1 10.0 NaN NaN NaN
6 1 2.0 NaN NaN NaN
7 2 32.0 100.0 32.0 100.0
8 2 32.0 242.0 NaN NaN
9 2 3.0 NaN NaN NaN
10 3 2.0 NaN 23.0 100.0
11 3 23.0 100.0 NaN NaN
12 3 NaN NaN NaN NaN
13 3 23.0 NaN NaN NaN
Converting missing values to empty strings is possible, but get mixed values - numeric and strings, so next processing should be problematic:
df['A_MAX'] = df.groupby('Serial')['A'].transform('max')
df['A_MAX'] = df['A_MAX'].mask(df['Serial'].duplicated(), '')
s = (df.sort_values(['Serial','A'], ascending=[True, False])
.dropna(subset=['B'])
.drop_duplicates('Serial')
.set_index('Serial')['B'])
df['B_corresponding'] = df['Serial'].map(s).mask(df['Serial'].duplicated(), '')
print (df)
Serial A B A_MAX B_corresponding
0 1 12.0 NaN 31 203
1 1 31.0 NaN
2 1 NaN NaN
3 1 12.0 NaN
4 1 31.0 203.0
5 1 10.0 NaN
6 1 2.0 NaN
7 2 32.0 100.0 32 100
8 2 32.0 242.0
9 2 3.0 NaN
10 3 2.0 NaN 23 100
11 3 23.0 100.0
12 3 NaN NaN
13 3 23.0 NaN
You could also use dictionaries to achieve the same if you are not so inclined to only use pandas.
a_to_b_mapping = df.groupby('A')['B'].min().to_dict()
series_to_a_mapping = df.groupby('Series')['A'].max().to_dict()
agg_df = {}
for series, a in series_to_a_mapping.items():
agg_df.append((series, a, a_to_b_mapping.get(a, None)))
agg_df = pd.DataFrame(agg_df, columns=['Series', 'A_max', 'B_corresponding'])
agg_df.head()
Series A_max B_corresponding
0 1 31.0 203.0
1 2 32.0 100.0
2 3 23.0 100.0
If you want, you could join this to original dataframe and mask duplicates.
dft = df.join(final_df.set_index('Serial'), on='Serial', how='left')
dft['A_max'] = dft['A_max'].mask(dft['A_max'].duplicated(), '')
dft['B_corresponding'] = dft['B_corresponding'].mask(dft['B_corresponding'].duplicated(), '')
dft

concat result of groupby pandas

I am raising this question for learning a new method for myself.
I have a dataframe like below,
ID Value
0 1 10
1 1 12
2 1 14
3 1 16
4 1 18
5 2 32
6 2 12
7 2 -8
8 2 -28
9 2 -48
10 2 -68
11 3 12
12 3 1
13 3 43
I want to convert this into:
ID Value ID Value ID Value
0 1.0 10.0 2 32 3.0 12.0
1 1.0 12.0 2 12 3.0 1.0
2 1.0 14.0 2 -8 3.0 43.0
3 1.0 16.0 2 -28 NaN NaN
4 1.0 18.0 2 -48 NaN NaN
5 NaN NaN 2 -68 NaN NaN
one way to solve this,
print
pd.concat([df[df['ID']==1].reset_index(drop=True),df[df['ID']==2].reset_index(drop=True),df[df['ID']==3].reset_index(drop=True)],axis=1)
But I'm thinking can I do the same concat operation for each groupby method result instead of filtering by value?
Any better/new approaches are more appreciated.
Thanks in advance.
Yup, very possible and quite simple with pd.concat, in fact.
df = pd.concat({k : g.reset_index(drop=True) for k, g in df.groupby('ID')}, axis=1)
df.columns = df.columns.droplevel(0)
Or, a minor variation in Dark's (now deleted) answer (which does not give you the opportunity to specify column suffixes automatically) -
pd.concat([g.reset_index(drop=True) for _, g in df.groupby('ID')], axis=1)
df
ID Value ID Value ID Value
0 1.0 10.0 2 32 3.0 12.0
1 1.0 12.0 2 12 3.0 1.0
2 1.0 14.0 2 -8 3.0 43.0
3 1.0 16.0 2 -28 NaN NaN
4 1.0 18.0 2 -48 NaN NaN
5 NaN NaN 2 -68 NaN NaN
Those column names are terrible, though. Rather than dropping the first level, you should consider concatenating them to form a pre/suf-fix for the second level. That should be a good exercise for you with df.columns.map.

How to swap index and values on pandas dataframe

I have some data, in which the index is a threshold, and the values are trns (true negative rates) for two classes, 0 and 1.
I want to get a dataframe, indexed by the tnr, of the threshold that corresponds to that tnr, for each class. Essentially, I want this:
I am able to achieve this effect by using the following:
pd.concat([pd.Series(data[0].index.values, index=data[0]),
pd.Series(data[1].index.values, index=data[1])],
axis=1)
Or, generalizing to any number of columns:
def invert_dataframe(df):
return pd.concat([pd.Series(df[col].index.values,
index=df[col]) for col in df.columns],
axis=1)
However, this seems extremely hacky and error prone. Is there a better way to do this, and is there maybe native Pandas functionality that would do this?
You can use stack with pivot:
data = pd.DataFrame({0:[10,20,31],10:[4,22,36],
1:[7,5,6]}, index=[2.1,1.07,2.13])
print (data)
0 1 10
2.10 10 7 4
1.07 20 5 22
2.13 31 6 36
df = data.stack().reset_index()
df.columns = list('abc')
df = df.pivot(index='c', columns='b', values='a')
print (df)
b 0 1 10
c
4 NaN NaN 2.10
5 NaN 1.07 NaN
6 NaN 2.13 NaN
7 NaN 2.10 NaN
10 2.10 NaN NaN
20 1.07 NaN NaN
22 NaN NaN 1.07
31 2.13 NaN NaN
36 NaN NaN 2.13

How to get indexes of values in a Pandas DataFrame?

I am sure there must be a very simple solution to this problem, but I am failing to find it (and browsing through previously asked questions, I didn't find the answer I wanted or didn't understand it).
I have a dataframe similar to this (just much bigger, with many more rows and columns):
x val1 val2 val3
0 0.0 10.0 NaN NaN
1 0.5 10.5 NaN NaN
2 1.0 11.0 NaN NaN
3 1.5 11.5 NaN 11.60
4 2.0 12.0 NaN 12.08
5 2.5 12.5 12.2 12.56
6 3.0 13.0 19.8 13.04
7 3.5 13.5 13.3 13.52
8 4.0 14.0 19.8 14.00
9 4.5 14.5 14.4 14.48
10 5.0 15.0 19.8 14.96
11 5.5 15.5 15.5 15.44
12 6.0 16.0 19.8 15.92
13 6.5 16.5 16.6 16.40
14 7.0 17.0 19.8 18.00
15 7.5 17.5 17.7 NaN
16 8.0 18.0 19.8 NaN
17 8.5 18.5 18.8 NaN
18 9.0 19.0 19.8 NaN
19 9.5 19.5 19.9 NaN
20 10.0 20.0 19.8 NaN
In the next step, I need to compute the derivative dVal/dx for each of the value columns (in reality I have more than 3 columns, so I need to have a robust solution in a loop, I can't select the rows manually each time). But because of the NaN values in some of the columns, I am facing the problem that x and val are not of the same dimension. I feel the way to overcome this would be to only select only those x intervals, for which the val is notnull. But I am not able to do that. I am probably making some very stupid mistakes (I am not a programmer and I am very untalented, so please be patient with me:) ).
Here is the code so far (now that I think of it, I may have introduced some mistakes just by leaving some old pieces of code because I've been messing with it for a while, trying different things):
import pandas as pd
import numpy as np
df = pd.read_csv('H:/DocumentsRedir/pokus/dataframe.csv', delimiter=',')
vals = list(df.columns.values)[1:]
for i in vals:
V = np.asarray(pd.notnull(df[i]))
mask = pd.notnull(df[i])
X = np.asarray(df.loc[mask]['x'])
derivative=np.diff(V)/np.diff(X)
But I am getting this error:
ValueError: operands could not be broadcast together with shapes (20,) (15,)
So, apparently, it did not select only the notnull values...
Is there an obvious mistake that I am making or a different approach that I should adopt? Thanks!
(And another less important question: is np.diff the right function to use here or had I better calculated it manually by finite differences? I'm not finding numpy documentation very helpful.)
To calculate dVal/dX:
dVal = df.iloc[:, 1:].diff() # `x` is in column 0.
dX = df['x'].diff()
>>> dVal.apply(lambda series: series / dX)
val1 val2 val3
0 NaN NaN NaN
1 1 NaN NaN
2 1 NaN NaN
3 1 NaN NaN
4 1 NaN 0.96
5 1 NaN 0.96
6 1 15.2 0.96
7 1 -13.0 0.96
8 1 13.0 0.96
9 1 -10.8 0.96
10 1 10.8 0.96
11 1 -8.6 0.96
12 1 8.6 0.96
13 1 -6.4 0.96
14 1 6.4 3.20
15 1 -4.2 NaN
16 1 4.2 NaN
17 1 -2.0 NaN
18 1 2.0 NaN
19 1 0.2 NaN
20 1 -0.2 NaN
We difference all columns (except the first one), and then apply a lambda function to each column which divides it by the difference in column X.

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