I have the data set 'datatoread' and following code as below, populates the dataframe 'MyData'.
datatoread <- "X Y
29 21
18 23
28 24
16 26
3 27
18 29
2 33
3 37
26 39
2 42
25 47
9 54
13 57
17 58
29 60
5 63
23 66
4 69
3 72
17 73
7 73
12 72
8 69
20 66
12 63
8 60
28 58
3 57
18 54
11 47
21 42
8 39
1 37
16 29
3 27
17 22
3 19
6 17
19 14
18 10"
MyData <- read.table(textConnection(datatoread), header = TRUE)
closeAllConnections()
MyData
What I want to do is:
Plot the data, find the mid value on X-axis and draw a vertical line from that X-value until Y-value (the corresponding Y coordinate to that mid-X).
Segment the left half to the mid-X into equal distance (as shown in the picture below) and tabulate the segments in a way that results, say,
Result (please note these are only indicative for coordinates of segments, actual might differ)
Seg X1 Y1 X2 Y2
seg1 18 23 29 21
seg2 29 21 28 24
. . . . .
For first part of it, I tried in SAS as:
data Trapezoidal;
set x_y end=last;
retain integral;
lag_x=lag(x); lag_y = lag(y);
if _N_ eq 1 then integral = 0;
else integral = integral + (x - lag_x) * (y + lag_y) / 2;
run;
What could be equivalent code in R or SQL?
Alternatively, assume this data:
x <- seq(-12,12,by=1)
y <- dnorm(x,mean=2.5,sd=5)
plot(x,y, type = "l")
z <- cbind(x,y)
plot(z, type = "l")
You can work on 'z' dataframe as well.
Related
df = pd.DataFrame(np.random.randint(0,100,size=(15, 3)), columns=list('NMO'))
df['Catgeory1'] = ['I','I','I','I','I','G','G','G','G','G','P','P','I','I','P']
df['Catgeory2'] = ['W','W','C','C','C','W','W','W','W','W','O','O','O','O','O']
Imagining this df is much larger with many more categories, how might I sort the list, retaining all the characteristics of any given row, by a determined order. Ex. Sorting the df only by 'category1', such that all the P's are first, the I's, then G's.
You can use categorical type:
cat_type = pd.CategoricalDtype(categories=["P", "I", "G"], ordered=True)
df['Category1'] = df['Category1'].astype(cat_type)
print(df.sort_values(by='Category1'))
Prints:
N M O Category1 Category2
10 49 37 44 P O
11 72 64 66 P O
14 39 98 32 P O
0 93 12 89 I W
1 20 74 21 I W
2 25 22 24 I C
3 47 11 33 I C
4 60 16 34 I C
12 0 90 6 I O
13 13 35 80 I O
5 84 64 67 G W
6 70 47 83 G W
7 61 57 76 G W
8 19 8 3 G W
9 7 8 5 G W
For PIG order (reverse alphabetical order):
df.sort_values('Catgeory1',ascending=False)
For custom sorting:
df['Catgeory1'] = pd.Categorical(df['Catgeory1'], ['P','G','I'])
df = df.sort_values('Catgeory1')
Let's assume we have a 3D Tensor of shape a = [batch_size, length, 1] and we want to discard every 5th sample from the length axis. The new indices for every batch element could be calculated as indices = tf.where(tf.range(a.shape[1]) % 5 != 0).
Could you please help me with an operation that obtains the shorter Tensor shape b = [batch_size, length2, 1], where length2 = 4/5 * length ? I assume this is attainable with tf.gather_nd, but I am having an issue with providing the indices in the right format. It does not simply work to tile the indices Tensor batch_size times and provide the resulting 2D tensor to tf.gather_nd taking the 3D tensor as parameters.
Thank you.
You can simply do the following:
import tensorflow as tf
# Example data
a = tf.reshape(tf.range(60), [5, 12, 1])
print(a.numpy()[:, :, 0])
# [[ 0 1 2 3 4 5 6 7 8 9 10 11]
# [12 13 14 15 16 17 18 19 20 21 22 23]
# [24 25 26 27 28 29 30 31 32 33 34 35]
# [36 37 38 39 40 41 42 43 44 45 46 47]
# [48 49 50 51 52 53 54 55 56 57 58 59]]
# Mask every one in five items
mask = tf.not_equal(tf.range(tf.shape(a)[1]) % 5, 0)
b = tf.boolean_mask(a, mask, axis=1)
# Show result
print(b.numpy()[:, :, 0])
# [[ 1 2 3 4 6 7 8 9 11]
# [13 14 15 16 18 19 20 21 23]
# [25 26 27 28 30 31 32 33 35]
# [37 38 39 40 42 43 44 45 47]
# [49 50 51 52 54 55 56 57 59]]
I have the following data frame, A, and would like to make a violin/box plot of the last data points (or any other selected) for all IDs in a time series, i.e. for time=90 the values for ID = 1...10 should be plotted.
A = data.frame(ID = rep(seq(1,5),each=10),
time = rep(seq(0,90,by = 10),5),
value = rnorm(50))
ID time value
1 1 0 0.056152116
2 1 10 0.560673698
3 1 20 -0.240922725
4 1 30 -1.054686869
5 1 40 -0.734477812
6 1 50 1.123602646
7 1 60 -2.242830898
8 1 70 -0.818526167
9 1 80 1.476234401
10 1 90 -0.332324134
11 2 0 -1.486034438
12 2 10 0.222252053
13 2 20 -0.675720560
14 2 30 -3.144918043
15 2 40 3.058383376
16 2 50 0.978174555
17 2 60 -0.280927730
18 2 70 -0.188338714
19 2 80 -1.115583389
20 2 90 0.362044729
...
41 5 0 0.687402844
42 5 10 -1.127714642
43 5 20 0.117758547
44 5 30 0.507666153
45 5 40 0.205580300
46 5 50 -1.033018214
47 5 60 -1.906279605
48 5 70 0.117539035
49 5 80 -0.968888556
50 5 90 0.122049005
Try this:
set.seed(42)
A = data.frame(ID = rep(seq(1,5),each=10),
time = rep(seq(0,90,by = 10),5),
value = rnorm(50))
library(ggplot2)
library(dplyr)
filter(A, time == 90) %>%
ggplot(aes(y = value)) +
geom_boxplot()
Created on 2020-06-09 by the reprex package (v0.3.0)
I have my data as
data = pd.DataFrame({'A':[3,50,50,60],'B':[49,5,37,59],'C':[15,34,43,6],'D':[35,39,10,25]})
If I use cut this way
p = ['A','S','T','U','V','C','Z']
bins = [0,30,35,40,45,50,55,60]
data['A*'] = pd.cut(data.A,bins,labels=p)
print(data)
I get
A B C D A*
0 3 49 15 35 A
1 50 5 34 39 V
2 50 37 43 10 V
3 60 59 6 25 Z
How would I cut it to get
A B C D A*
0 3 49 15 35 3A
1 50 5 34 39 50V
2 50 37 43 10 50V
3 60 59 6 25 60Z
I tried this but doesn't work
for x in data.A:
p = [str(x)+'A',str(x)+'S',str(x)+'T',str(x)+'U',str(x)+'V',str(x)+'C',str(x)+'Z']
bins = [0,30,35,40,45,50,55,60]
It gives me this
A B C D A*
0 3 49 15 35 60A
1 50 5 34 39 60V
2 50 37 43 10 60V
3 60 59 6 25 60Z
Convert column A to strings and categoricals from pd.cut too and join together:
p = ['A','S','T','U','V','C','Z']
bins = [0,30,35,40,45,50,55,60]
data['A*'] = data.A.astype(str) + pd.cut(data.A,bins,labels=p).astype(str)
print(data)
A B C D A*
0 3 49 15 35 3A
1 50 5 34 39 50V
2 50 37 43 10 50V
3 60 59 6 25 60Z
EDIT:
For processing all columns is possible use DataFrame.apply:
data = data.apply(lambda x: x.astype(str) + pd.cut(x,bins,labels=p).astype(str))
print(data)
A B C D
0 3A 49V 15A 35S
1 50V 5A 34S 39T
2 50V 37T 43U 10A
3 60Z 59Z 6A 25A
I lose my multiIndex when I try to pd.concat a second subtotal. I'm able to add the first subtotal but not the second which is a sum of B0.
This is how my current df is:
lvl0 a b
lvl1 bar foo bah foo
A0 B0 C0 D0 1 0 3 2
D1 5 4 7 6
First Total 6 4 10 8
C1 D0 9 8 11 10
D1 13 12 15 14
First Total 22 20 26 24
C2 D0 17 16 19 18
After trying to add the second subtotal I get this:
lvl0 a b
lvl1 bar foo bah foo
(A0, B0, C2, First Total) 38 36 42 40
(A0, B0, C3, D0) 25 24 27 26
(A0, B0, C3, D1) 29 28 31 30
(A0, B0, C3, First Total) 54 52 58 56
(A0, B0, Second Total) 120 112 136 128
(A0, B1, C0, D0) 33 32 35 34
(A0, B1, C0, D1) 37 36 39 38
(A0, B1, C0, First Total) 70 68 74 72
(A0, B1, C1, D0) 41 40 43 42
You should be able to copy and paste the code below to test
import pandas as pd
import numpy as np
# creating multiIndex
def mklbl(prefix, n):
return ["%s%s" % (prefix, i) for i in range(n)]
miindex = pd.MultiIndex.from_product([mklbl('A', 4),
mklbl('B', 2),
mklbl('C', 4),
mklbl('D', 2)])
micolumns = pd.MultiIndex.from_tuples([('a', 'foo'), ('a', 'bar'),
('b', 'foo'), ('b', 'bah')],
names=['lvl0', 'lvl1'])
dfmi = pd.DataFrame(np.arange(len(miindex) * len(micolumns))
.reshape((len(miindex), len(micolumns))),
index=miindex,
columns=micolumns).sort_index().sort_index(axis=1)
# My code STARTS HERE
# creating the first subtotal
print(dfmi.index)
df1 = dfmi.groupby(level=[0,1,2]).sum()
df2 = dfmi.groupby(level=[0, 1]).sum()
df1 = df1.set_index(np.array(['First Total'] * len(df1)), append=True)
dfmi = pd.concat([dfmi, df1]).sort_index(level=[0, 1])
print(dfmi)
# this is where the multiIndex is lost
df2 = df2.set_index(np.array(['Second Total'] * len(df2)), append=True)
dfmi = pd.concat([dfmi, df2]).sort_index(level=[1])
print(dfmi)
How I would want it to look:
lvl0 a b
lvl1 bar foo bah foo
A0 B0 C0 D0 1 0 3 2
D1 5 4 7 6
First Total 6 4 10 8
C1 D0 9 8 11 10
D1 13 12 15 14
First Total 22 20 26 24
C2 D0 17 16 19 18
D1 21 20 23 22
First Total 38 36 42 40
C3 D0 25 24 27 26
D1 29 28 31 30
First Total 54 52 58 56
Second Total 120 112 136 128
B1 C0 D0 33 32 35 34
D1 37 36 39 38
First Total 70 68 74 72
C1 D0 41 40 43 42
D1 45 44 47 46
First Total 86 84 90 88
C2 D0 49 48 51 50
D1 53 52 55 54
First Total 102 100 106 104
C3 D0 57 56 59 58
D1 61 60 63 62
First Total 118 116 122 120
Second Total 376 368 392 384
the first total is sum of level 2,
the second total is sum of level 1
dfmi has a 4-level MultiIndex:
In [208]: dfmi.index.nlevels
Out[208]: 4
df2 has a 3-level MultiIndex. Instead, if you use
df2 = df2.set_index([np.array(['Second Total'] * len(df2)), [''] * len(df2)], append=True)
then df2 ends up with a 4-level MultiIndex. When dfmi and df2 have the same number of levels,
then pd.concat([dfmi, df2]) produces the desired result.
One problem you may face when sorting by index labels is that it relies on the strings 'First' and 'Second'
appearing last in alphabetic order. An alterative to sorting by index would be assigning a numeric order column
and sorting by that instead:
dfmi['order'] = range(len(dfmi))
df1['order'] = dfmi.groupby(level=[0,1,2])['order'].last() + 0.1
df2['order'] = dfmi.groupby(level=[0,1])['order'].last() + 0.2
...
dfmi = pd.concat([dfmi, df1, df2])
dfmi = dfmi.sort_values(by='order')
Incorporating Scott Boston's improvement, the code would then look like this:
import pandas as pd
import numpy as np
def mklbl(prefix, n):
return ["%s%s" % (prefix, i) for i in range(n)]
miindex = pd.MultiIndex.from_product([mklbl('A', 4),
mklbl('B', 2),
mklbl('C', 4),
mklbl('Z', 2)])
micolumns = pd.MultiIndex.from_tuples([('a', 'foo'), ('a', 'bar'),
('b', 'foo'), ('b', 'bah')],
names=['lvl0', 'lvl1'])
dfmi = pd.DataFrame(np.arange(len(miindex) * len(micolumns))
.reshape((len(miindex), len(micolumns))),
index=miindex,
columns=micolumns).sort_index().sort_index(axis=1)
df1 = dfmi.groupby(level=[0,1,2]).sum()
df2 = dfmi.groupby(level=[0, 1]).sum()
dfmi['order'] = range(len(dfmi))
df1['order'] = dfmi.groupby(level=[0,1,2])['order'].last() + 0.1
df2['order'] = dfmi.groupby(level=[0,1])['order'].last() + 0.2
df1 = df1.assign(lev4='First').set_index('lev4', append=True)
df2 = df2.assign(lev3='Second', lev4='').set_index(['lev3','lev4'], append=True)
dfmi = pd.concat([dfmi, df1, df2])
dfmi = dfmi.sort_values(by='order')
dfmi = dfmi.drop(['order'], axis=1)
print(dfmi)
which yields
lvl0 a b
lvl1 bar foo bah foo
A0 B0 C0 Z0 1 0 3 2
Z1 5 4 7 6
First 6 4 10 8
C1 Z0 9 8 11 10
Z1 13 12 15 14
First 22 20 26 24
C2 Z0 17 16 19 18
Z1 21 20 23 22
First 38 36 42 40
C3 Z0 25 24 27 26
Z1 29 28 31 30
First 54 52 58 56
Second 120 112 136 128
...
#unutbu points out the nature of the problem. df2 has three levels of a multiindex and you need a 4th level.
I would use assign and set_index to create that fourth level:
df2 = df2.assign(lev3='Second Total', lev4='').set_index(['lev3','lev4'], append=True)
This avoids calculating the length of the dataframe.