I need to create a text file called Names_ages.txt using the following list
Josh 14
Lauren 16
Paul 18
Mason 15
Lily 17
Stephanie 15
If anyone can help me and understand how to create one, that would be greatly appreciated :)
This creates a text file from a list:
outfile = open("Names_ages.txt",'w')
name_list = ["Josh 14", "Lauren 16", "Paul 18", "Mason 15", "Lily 17", "Stephanie 15"]
outfile.write("\n".join(name_list))
Related
I´m just running the following example from GGEBiplotGUI package and of course, it works properly.
library(GGEBiplotGUI)
data("Ontario")
Ontario
GGEBiplot(Data = Ontario)
But when I download "Ontario" data and I want to run the above cited script on my PC. See the example below.
Ontario <- read.csv("Book.csv")
library(GGEBiplotGUI)
GGEBiplot(Data = Ontario)
The result is the following table (from column 0 to 10) taking numbers (From 1 to 17) as genotypes and "X" as another location.
See the result below please.
X BH93 EA93 HW93 ID93 KE93 NN93 OA93 RN93 WP93
1 ann 4.460 4.150 2.849 3.084 5.940 4.450 4.351 4.039 2.672
2 ari 4.417 4.771 2.912 3.506 5.699 5.152 4.956 4.386 2.938
3 aug 4.669 4.578 3.098 3.460 6.070 5.025 4.730 3.900 2.621
4 cas 4.732 4.745 3.375 3.904 6.224 5.340 4.226 4.893 3.451
5 del 4.390 4.603 3.511 3.848 5.773 5.421 5.147 4.098 2.832
6 dia 5.178 4.475 2.990 3.774 6.583 5.045 3.985 4.271 2.776
7 ena 3.375 4.175 2.741 3.157 5.342 4.267 4.162 4.063 2.032
8 fun 4.852 4.664 4.425 3.952 5.536 5.832 4.168 5.060 3.574
9 ham 5.038 4.741 3.508 3.437 5.960 4.859 4.977 4.514 2.859
10 har 5.195 4.662 3.596 3.759 5.937 5.345 3.895 4.450 3.300
11 kar 4.293 4.530 2.760 3.422 6.142 5.250 4.856 4.137 3.149
12 kat 3.151 3.040 2.388 2.350 4.229 4.257 3.384 4.071 2.103
13 luc 4.104 3.878 2.302 3.718 4.555 5.149 2.596 4.956 2.886
14 m12 3.340 3.854 2.419 2.783 4.629 5.090 3.281 3.918 2.561
15 reb 4.375 4.701 3.655 3.592 6.189 5.141 3.933 4.208 2.925
16 ron 4.940 4.698 2.950 3.898 6.063 5.326 4.302 4.299 3.031
17 rub 3.786 4.969 3.379 3.353 4.774 5.304 4.322 4.858 3.382
How can I fix this problem? I mean, in order to avoid "rownames" and "x" as a variables in the GGEBiplotGUI analysis.
I have also tried with these codes and they didn´t work:
attributes(Ontario)$row.names <- NULL
print(Ontario, row.names = F)
row.names(Ontario) <- NULL
Ontario[, -1] ## It deletes the first column not the 0 one.
Many thanks in advance!
This code worked properly.
Ontario <- read.csv("Libro.csv")
rownames(Ontario)<-Ontario$X
Ontario1<-Ontario[,-1]
library(GGEBiplotGUI)
GGEBiplot(Data = Ontario)
I am trying to convert the string to the type of 'datetime' in python. My data match the format, but still get the
'ValueError: time data 11 11 doesn't match format specified'
I am not sure where does the "11 11" in the error come from.
My code is
train_df['date_captured1'] = pd.to_datetime(train_df['date_captured'], format="%Y-%m-%d %H:%M:%S")
Head of data is
print (train_df.date_captured.head())
0 2011-05-13 23:43:18
1 2012-03-17 03:48:44
2 2014-05-11 11:56:46
3 2013-10-06 02:00:00
4 2011-07-12 13:11:16
Name: date_captured, dtype: object
I tried the following by just selecting the first string and running the code with same datetime format. They all work without problem.
dt=train_df['date_captured']
dt1=dt[0]
date = datetime.datetime.strptime(dt1, "%Y-%m-%d %H:%M:%S")
print(date)
2011-05-13 23:43:18
and
dt1=pd.to_datetime(dt1, format='%Y-%m-%d %H:%M:%S')
print (dt1)
2011-05-13 23:43:18
But why wen I using the same format in pd.to_datetime to convert all the data in the column, it comes up with the error above?
Thank you.
I solved it.
train_df['date_time'] = pd.to_datetime(train_df['date_captured'], errors='coerce')
print (train_df[train_df.date_time.isnull()])
I found in line 100372, the date_captured value is '11 11'
category_id date_captured ... height date_time
100372 10 11 11 ... 747 NaT
So the code with errors='coerce' will replace the invalid parsing with NaN.
Thank you.
I was wondering if you could help me to convert the data format per below:
01/01/2018 to Jan 18
09/30/2018 to Q3 18
=Table.AddColumn(#"Removed Columns2", "Custom", each Date.ToText([Report Date],"MMM")&"-"&Date.Year([Report Date],"YY"))
This is what I have tried so far and no results:
Thank you very much for your help!
You're nearly there.
Try
Custom1 = Table.AddColumn(#"Removed Columns2", "MMM-YY", each Date.ToText([Report Date],"MMM-yy")),
Custom2 = Table.AddColumn(#"Custom1", "QYY", each Number.ToText(Date.QuarterOfYear([Report Date])) & Date.ToText([Report Date], "yy"))
I have a spark dataFrame that looks like this:
id dates value
1 11 2013-11-15 10
2 11 2013-11-16 15
3 22 2013-11-15 20
4 22 2013-11-16 21
5 22 2013-11-17 3
I wish to retain the value from the previous date per id.
The final result should look like this:
id dates value prev_value
1 11 2013-11-15 10 NA
2 11 2013-11-16 15 10
3 22 2013-11-15 20 NA
4 22 2013-11-16 21 20
5 22 2013-11-17 3 21
The solution from this question would not work for various reasons.
I would appreciate the help!
So after playing with it for a while, here's the workaround that I found:
First of all, here's the example DF
id<-c(11,11,22,22,22)
dates<-as.Date(c('2013-11-15','2013-11-16','2013-11-15','2013-11-16','2013-11-17'), "%Y-%m-%d")
value <- c(10,15,20,21,3)
example<-as.DataFrame(data.frame(id=id,dates=dates, value))
I copy the example DF and add 1 day to the original date, then rename the column
example_p <- example
example_p$dates <- date_add(example_p$dates, 1)
colnames(example_p) <- c("id", "dates", "prev_value")
Finally, I merge the new DF to the original one
result <- select(merge(example, example_p, by = intersect(names(example),names(example_p))
, all.x = T), c("id_x", "dates_x", "value", "prev_value"))
showDF(result)
+----+----------+-----+----------+
|id_x| dates_x|value|prev_value|
+----+----------+-----+----------+
|22.0|2013-11-15| 20.0| null|
|11.0|2013-11-15| 10.0| null|
|11.0|2013-11-16| 15.0| 10.0|
|22.0|2013-11-16| 21.0| 20.0|
|22.0|2013-11-17| 3.0| 21.0|
+----+----------+-----+----------+
Obviously, this is somehow clumsy and I will be happy to give the points to anyone who can suggest a solution that would work faster than this.
I am trying to pre-process some text using regex in ruby to input into a mapper job and would like to split on the carriage return denoting the paragraph.
The text will be coming into the mapper using ARGF.each as part of a hadoop streaming job
"\"Walter Elliot, born March 1, 1760, married, July 15, 1784, Elizabeth,\r\n"
"daughter of James Stevenson, Esq. of South Park, in the county of\r\n"
"Gloucester, by which lady (who died 1800) he has issue Elizabeth, born\r\n"
"June 1, 1785; Anne, born August 9, 1787; a still-born son, November 5,\r\n"
"1789\"\r\n"
"\r\n" # <----- this is where I would like to split
"Precisely such had the paragraph originally stood from the printer's\r\n"
Once I have done this I will chomp the newline /carriage return of each line.
This will look something like this:
ARGF.each do |text|
paragraph = text.split(INSERT_REGEX_HERE)
#some more blah will happen beyond here
end
UPDATE:
The desired output then is an array as follows:
[
[0] "\"Walter Elliot, born March 1, 1760, married, July 15, 1784, Elizabeth,\r\n"
"daughter of James Stevenson, Esq. of South Park, in the county of\r\n"
"Gloucester, by which lady (who died 1800) he has issue Elizabeth, born\r\n"
"June 1, 1785; Anne, born August 9, 1787; a still-born son, November 5,\r\n"
"1789\"\r\n"
[1] "Precisely such had the paragraph originally stood from the printer's\r\n"
]
Ultimately what I want is the following array with no carriage returns within the array:
[
[0] "\"Walter Elliot, born March 1, 1760, married, July 15, 1784, Elizabeth,"
"daughter of James Stevenson, Esq. of South Park, in the county of"
"Gloucester, by which lady (who died 1800) he has issue Elizabeth, born"
"June 1, 1785; Anne, born August 9, 1787; a still-born son, November 5,"
"1789\""
[1] "Precisely such had the paragraph originally stood from the printer's"
]
Thanks in advance for any insights.
Beware when you do ARGF.each do |text|, the text will be every single line, NOT the whole text block.
You can provide ARGF.each a special line separator, it will return you two "lines", which are the two paragraphs in your case.
Try this:
paragraphs = ARGF.each("\r\n\r\n").map{|p| p.gsub("\r\n","")}
First, split input into two paragraphs, then use gsub to remove unwanted line breaks.
To split the text use:
result = text.gsub(/(?<!\")\\r\\n|(?<=\\\")\\r\\n/, '').split(/[\r\n]+\"\\r\\n\".*?[\r\n]+/)