I have the following Dataframe called df_cam_cb_days:
3m 6m 9m 1y 18m 24m Effective
2021-03-30 49 49 49 49 49 49 2021-03-31
2021-05-13 40 44 44 44 44 44 2021-05-14
2021-06-08 0 26 26 26 26 26 2021-06-09
2021-07-14 0 36 36 36 36 36 2021-07-15
2021-08-31 0 26 48 48 48 48 2021-09-01
2021-10-13 0 0 43 43 43 43 2021-10-14
2021-12-14 0 0 27 62 62 62 2021-12-15
2022-01-26 0 0 0 43 43 43 2022-01-27
2022-03-30 0 0 0 14 63 63 2022-03-31
2022-05-11 0 0 0 0 42 42 2022-05-12
2022-06-08 0 0 0 0 28 28 2022-06-09
2022-07-13 0 0 0 0 35 35 2022-07-14
2022-08-31 0 0 0 0 27 49 2022-09-01
2022-10-12 0 0 0 0 0 42 2022-10-13
2022-12-14 0 0 0 0 0 63 2022-12-15
2023-01-25 0 0 0 0 0 42 2023-01-26
2023-02-10 0 0 0 0 0 15 2023-02-11
and I have the following function that receives the DataFrame and an array:
mon_policy =np.array([.5,
.75,
.75,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1])
#returns numpy array with Breakeven info
def cam_be_mon(mp,df):
columns = ['3m','6m','9m','1y','18m','24m']
days_array = np.array([0,0,0,0,0,0])
days_array = df_cam_cb_days[columns].sum(axis=0).values
data_array= df_cam_cb_days[columns].values.T
c= np.log(mp/36000+1)
be = np.dot(data_array,c)
be = (np.exp(be[0:])-1)*36000/days_array
return be
target = np.array([.3525,.415,.475,.56,.715,.916366])
cam_be_mon(mon_policy,df_cam_cb_days)
The Function as is returns the solution: array([0.61281788, 0.76943154, 0.84886388, 0.88890188,
0.92955637, 0.95151633])
I need to find the set of <mon_policy> that returns a solution = to target , or the closest if there is no solution.
I found the answer with scipy.optimize
Related
I've a Dataframe as below. (resulted from pivot_table() method)
Location Loc 2 Loc 3 Loc 5 Loc 8 Loc 9
Item
1 404 317 272 113 449
3 1,205 870 846 371 1,632
5 208 218 128 31 268
7 107 54 57 17 179
9 387 564 245 83 571
10 364 280 115 34 252
16 104 80 72 22 143
17 111 85 44 10 209
18 124 182 67 27 256
19 380 465 219 103 596
if you take a closer look at it, there are missing Locations (eg, Loc 1, Loc, 4, etc) and missing Items (eg, 2, 4,8, etc)
I want to export this to my Excel pre-defined Template which has all the Locations & Items & fill the table based on Items & Values.
I know I can export the dataframe to a different excel sheet & use SUMIFS() or INDEX(), MATCH() formulas. but, I want to do this directly from Python/Panda to excel.
Below should be the result after exporting
Loc 1 Loc 2 Loc 3 Loc 4 Loc 5 Loc 6 Loc 7 Loc 8 Loc 9
1 0 404 317 0 272 0 0 113 449
2 0 0 0 0 0 0 0 0 0
3 0 1205 870 0 846 0 0 371 1632
4 0 0 0 0 0 0 0 0 0
5 0 208 218 0 128 0 0 31 268
6 0 0 0 0 0 0 0 0 0
7 0 107 54 0 57 0 0 17 179
8 0 0 0 0 0 0 0 0 0
9 0 387 564 0 245 0 0 83 571
10 0 364 280 0 115 0 0 34 252
11 0 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0 0 0 0
13 0 0 0 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0
16 0 104 80 0 72 0 0 22 143
17 0 111 85 0 44 0 0 10 209
18 0 124 182 0 67 0 0 27 256
19 0 380 465 0 219 0 0 103 596
20 0 0 0 0 0 0 0 0 0
Use DataFrame.reindex with new index and columns values in arrays or lists:
idx = np.arange(1, 21)
cols = [f'Loc {x}' for x in np.arange(1, 10)]
df = df.reindex(index=idx, columns=cols, fill_value=0)
print (df)
Loc 1 Loc 2 Loc 3 Loc 4 Loc 5 Loc 6 Loc 7 Loc 8 Loc 9
1 0 404 317 0 272 0 0 113 449
2 0 0 0 0 0 0 0 0 0
3 0 1,205 870 0 846 0 0 371 1,632
4 0 0 0 0 0 0 0 0 0
5 0 208 218 0 128 0 0 31 268
6 0 0 0 0 0 0 0 0 0
7 0 107 54 0 57 0 0 17 179
8 0 0 0 0 0 0 0 0 0
9 0 387 564 0 245 0 0 83 571
10 0 364 280 0 115 0 0 34 252
11 0 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0 0 0 0
13 0 0 0 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0
16 0 104 80 0 72 0 0 22 143
17 0 111 85 0 44 0 0 10 209
18 0 124 182 0 67 0 0 27 256
19 0 380 465 0 219 0 0 103 596
20 0 0 0 0 0 0 0 0 0
I have the following glyph path:
<glyph glyph-name="right-nav-workflow" unicode="" d="M251 101c0 65 0 131 0 196 26 0 52 0 78 0 0 27 0 54 0 81-29 7-52 23-67 50-11 20-14 41-10 64 8 45 48 79 92 81 49 1 88-29 102-79 2 0 4 0 6 0 20 0 40 0 60 0 5 0 8 1 11 5 32 32 64 64 96 96 2 1 3 3 4 4 42-42 84-84 127-127-1 0-2-2-4-4-32-32-65-65-98-98-2-2-4-6-4-10 0-21 0-43 0-64 30-8 53-24 67-52 12-21 15-44 9-68-11-46-55-78-102-75-48 3-88 42-92 91-4 48 28 91 78 104 0 1 1 3 1 4 0 21 0 42 0 62 0 3-2 5-3 7-28 28-55 55-83 83-1 1-3 3-5 3-22 0-45 0-68 0-5-19-13-36-27-50-14-14-31-23-50-27 0-27 0-53 0-81 26 0 52 0 78 0 0-65 0-130 0-196-65 0-131 0-196 0z m157 156c-40 0-79 0-118 0 0-39 0-78 0-117 40 0 79 0 118 0 0 39 0 78 0 117z m275-58c0 33-26 59-59 59-32 0-59-26-59-59 0-33 27-59 60-59 32 0 58 26 58 59z m-334 216c33 0 59 27 59 59 0 33-26 59-59 59-33 0-59-26-59-59 0-33 26-59 59-59z m275-11c23 23 46 46 69 69-23 23-47 46-69 68-23-22-46-46-69-69 23-22 46-45 69-68z" horiz-adv-x="1000" />
How can I convert this into SVG? I tried to save it in a text file with svg extension but seems is not working.
Take the raw drawing data, put it in a path element, remove the glyph specific attributes, add an appropriate viewBox. Now you have something that works as an inline SVG. If you want to save it as a standalone SVG, then you need to add a name space declaration.
<svg height="400px" width="400px" viewBox="0 0 1000 1000">
<path transform="scale(1,-1) translate(0,-650)" fill="none" stroke="red" stroke-width="1" d="M251 101c0 65 0 131 0 196 26 0 52 0 78 0 0 27 0 54 0 81-29 7-52 23-67 50-11 20-14 41-10 64 8 45 48 79 92 81 49 1 88-29 102-79 2 0 4 0 6 0 20 0 40 0 60 0 5 0 8 1 11 5 32 32 64 64 96 96 2 1 3 3 4 4 42-42 84-84 127-127-1 0-2-2-4-4-32-32-65-65-98-98-2-2-4-6-4-10 0-21 0-43 0-64 30-8 53-24 67-52 12-21 15-44 9-68-11-46-55-78-102-75-48 3-88 42-92 91-4 48 28 91 78 104 0 1 1 3 1 4 0 21 0 42 0 62 0 3-2 5-3 7-28 28-55 55-83 83-1 1-3 3-5 3-22 0-45 0-68 0-5-19-13-36-27-50-14-14-31-23-50-27 0-27 0-53 0-81 26 0 52 0 78 0 0-65 0-130 0-196-65 0-131 0-196 0z m157 156c-40 0-79 0-118 0 0-39 0-78 0-117 40 0 79 0 118 0 0 39 0 78 0 117z m275-58c0 33-26 59-59 59-32 0-59-26-59-59 0-33 27-59 60-59 32 0 58 26 58 59z m-334 216c33 0 59 27 59 59 0 33-26 59-59 59-33 0-59-26-59-59 0-33 26-59 59-59z m275-11c23 23 46 46 69 69-23 23-47 46-69 68-23-22-46-46-69-69 23-22 46-45 69-68z"/>
</svg>
Update - Robert points out that the SVG Fonts spec has a y axis (0 on the bottom) that's inverted from the SVG norm (0 is the top) -> so you also need to flip the drawing along the x axis by using a scale (1,-1).
For the code:
dataset = pd.read_csv("/Users/Akshita/Desktop/EE660/donor_raw_data_medmean.csv", header=None, names=None)
# Separate data and label
X_label = dataset[1:19373][0]
X_data = dataset[1:19373]
print(X_data[X_label==1])
I get the output:(There are actually 4000~ samples with label=1)
0 1 2 3 4 5 6 7 8 9 ... 51 52 53 54 55 56 57 58 \
16386 1 17 60 0 1 0 0 0 0 1 ... 0 20 20 20 5 10 15 15
16396 1 137 60 0 1 0 0 0 0 1 ... 15 25 10 15 6 14 16 120
16399 1 89 54 0 1 0 0 0 0 1 ... 10 15 5 15 6 14 16 79
16402 1 89 75 0 1 0 0 0 0 1 ... 25 35 10 35 6 13 15 79
..
..
19356 1 101 80 1 0 0 1 0 0 2 ... 25 30 5 28 7 16 18 101
19363 1 65 70 1 0 0 1 0 0 1 ... 7 12 5 10 4 8 20 63
19372 1 29 70 0 0 0 1 0 0 2 ... 0 25 25 25 4 9 24 24
..
[859 rows x 61 columns]
and for
print(X_data[X_label==0])
I get the output:(There are about 15000~ samples with label=0)
0 1 2 3 4 5 6 7 8 9 ... 51 52 53 54 55 56 57 58 \
16384 0 17 74 0 1 0 0 0 0 1 ... 0 15 15 15 4 10 17 17
16385 0 17 60 0 1 0 0 0 0 2 ... 0 15 15 15 4 11 17 17
16387 0 29 67 0 1 0 0 0 0 1 ... 0 20 20 20 5 11 23 28
16388 0 53 60 0 1 0 0 0 0 1 ... 5 30 25 30 5 11 26 52
16389 0 65 49 0 1 0 0 0 0 1 ... 30 35 5 27 6 13 16 56
..
..
19369 0 137 77 1 0 1 0 0 0 1 ... 9 10 1 10 6 13 21 130
19370 0 29 60 1 0 0 1 0 0 1 ... 0 15 15 15 3 9 23 23
19371 0 129 78 1 0 0 1 0 0 2 ... 20 25 5 25 7 24 8 129
What can I be doing wrong?
I wrote two programs to run on Linux, each using a different algorithm, and I want to find a way (preferably using a benchmarking software) to compare the CPU usage and IO operations between these two programs.
Is there such a thing? and if yes, where can I find them. Thanks.
You can try hardinfo
Or there are like n different tools measuring system performance if measuring it while running your app solves your purpose
And you can also check this thread
You might try vmstat command:
vmstat 2 20 > vmstat.txt
20 samples of 2 seconds
bi = KB in, bo = KB out with wa = waiting for I/O
I/O can also increase cache demands
%CPU utilisation = us (user) = sy (system)
procs -----------memory---------- ---swap-- -----io---- -system-- ----cpu----
r b swpd free buff cache si so bi bo in cs us sy id wa
0 0 0 277504 17060 82732 0 0 91 87 1432 236 11 3 84 1
0 0 0 277372 17068 82732 0 0 0 24 1361 399 23 8 59 10
test start
0 1 0 275240 17068 82732 0 0 0 512 1342 305 24 4 69 4
2 1 0 275232 17068 82780 0 0 24 10752 4176 216 7 8 0 85
1 1 0 275240 17076 82732 0 0 12288 2590 5295 243 15 8 0 77
0 1 0 275240 17076 82748 0 0 8 11264 4329 214 6 12 0 82
0 1 0 275240 17076 82780 0 0 16 11264 4278 233 15 10 0 75
0 1 0 275240 17084 82780 0 0 19456 542 6563 255 10 7 0 83
0 1 0 275108 17084 82748 0 0 5128 3072 3501 265 16 37 0 47
3 1 0 275108 17084 82748 0 0 924 5120 8369 3845 12 33 0 55
0 1 0 275116 17092 82748 0 0 1576 85 11483 6645 5 50 0 45
1 1 0 275116 17092 82748 0 0 0 136 2304 689 3 9 0 88
2 1 0 275084 17100 82732 0 0 0 352 2374 800 14 26 0 61
0 0 0 275076 17100 82732 0 0 546 118 2408 1014 35 17 47 1
0 1 0 275076 17104 82732 0 0 0 62 1324 76 3 2 89 7
1 1 0 275076 17108 82732 0 0 0 452 1879 442 8 13 66 12
0 0 0 275116 17108 82732 0 0 800 352 2456 1195 19 17 56 8
0 1 0 275116 17112 82732 0 0 0 54 1325 76 4 1 88 8
test end
1 1 0 275116 17116 82732 0 0 0 510 1717 286 6 10 72 11
1 0 0 275076 17116 82732 0 0 1600 1152 3087 1344 23 29 41 7
I would like to convert this vector with strings in each row and spaces separating the elements within one string:
> v.input_red
[1] "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
[2] "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
[3] "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
to a dataframe with a column for each element. But I'm not quite sure how to extract the elements from the strings. Best way would be to convert the whole thing at once somehow, I guess..
Wanted result-dataframe (created manually):
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33 V34 V35
1 pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249
2 pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249
3 pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249
Thanks in advance!
Matthias
For quite some time, read.table and family have had a text argument that lets you read directly from character vectors. There's no need to write the object to a file first.
Your sample data...
v.input_red <- c("pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 ",
"pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 ",
"pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 ")
... directly read in:
read.table(text = v.input_red, header = FALSE)
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17
# 1 pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0
# 2 pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0
# 3 pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0
# V18 V19 V20 V21 V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33
# 1 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff
# 2 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff
# 3 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff
# V34 V35 V36 V37
# 1 0 249 0 0
# 2 0 249 0 0
# 3 0 249 0 0
Assuming file is a file name that you save on your system:
writeLines(v.input_red, file)
data <- read.table(file)
Is this solution what you were looking for?
s1 <- "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
s2 <- "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
s3 <- "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
df <- t(data.frame(strsplit(s1, " "),strsplit(s2, " "),strsplit(s3, " ")))
row.names(df) <- c("s1", "s2", "s3")
strsplit splits the string at each space char. Concatenated as data.frame gives you a df wih 3 columns. So you have to transpose it with t. I changes row names for better readability.