bigList = [list1,list2,list3]
I need a way to get the name of list1 using the zero index
bigList[0] just gives you all the items in the list
My original code prints all the items in the lists while the modified one says the index is out of range. I want it to give me the ilness names not all the symptoms.
Original code
https://i.stack.imgur.com/HDlA9.png
Modified
https://i.stack.imgur.com/5Qdh2.png
Related
I have a long list of strings (the main list), and a short list of strings. The short list is an exclude list, and I want to remove all occurences of the elements in the exclude list from the main list.
I've found these two ways of doing it, but none of them seem to work:
val fnrliste: MutableList<String> = ArrayList()
val excludeList = listOf("28030140259", "12050101833", "21089233132", "12050101833")
//Alternative 1:
for (fnr in excludeList) { fnrliste.remove(fnr) }
//Alternative 2:
fnrliste.removeAll(excludeList)
With both alternatives, the strings in the entries are still there when I list the contents of fnrliste.
As can be seen in this screenshot, some entries have been removed (different result with the two methods), but the first entry in the exclude list is still present:
What am I missing here?
Looks like your fnrliste contains some duplicates, so that's why you get different result. Also, as far as I can remember your screenshot, in case of removeAll the first element is "28030140259 " and not "28030140259" so everything is correct
I am a beginner in python and have encountered the following problem: I have a long list of strings (I took 3 now for the example):
ENSEMBL_IDs = ['ENSG00000040608',
'ENSG00000070371',
'ENSG00000070413']
which are partial matches of the data in column 0 of my DataFrame genes_df (first 3 entries shown):
genes_list = (['ENSG00000040608.28', 'RTN4R'],
['ENSG00000070371.91', 'CLTCL1'],
['ENSG00000070413.17', 'DGCR2'])
genes_df = pd.DataFrame(genes_list)
The task I want to perform is conceptually not that difficult: I want to compare each element of ENSEMBL_IDs to genes_df.iloc[:,0] (which are partial matches: each element of ENSEMBL_IDs is contained within column 0 of genes_df, as outlined above). If the element of EMSEMBL_IDs matches the element in genes_df.iloc[:,0] (which it does, apart from the extra numbers after the period ".XX" ), I want to return the "corresponding" value that is stored in the first column of the genes_df Dataframe: the actual gene name, 'RTN4R' as an example.
I want to store these in a list. So, in the end, I would be left with a list like follows:
`genenames = ['RTN4R', 'CLTCL1', 'DGCR2']`
Some info that might be helpful: all of the entries in ENSEMBL_IDs are unique, and all of them are for sure contained in column 0 of genes_df.
I think I am looking for something along the lines of:
`genenames = []
for i in ENSEMBL_IDs:
if i in genes_df.iloc[:,0]:
genenames.append(# corresponding value in genes_df.iloc[:,1])`
I am sorry if the question has been asked before; I kept looking and was not able to find a solution that was applicable to my problem.
Thank you for your help!
Thanks also for the edit, English is not my first language, so the improvements were insightful.
You can get rid of the part after the dot (with str.extract or str.replace) before matching the values with isin:
m = genes_df[0].str.extract('([^.]+)', expand=False).isin(ENSEMBL_IDs)
# or
m = genes_df[0].str.replace('\..*$', '', regex=True).isin(ENSEMBL_IDs)
out = genes_df.loc[m, 1].tolist()
Or use a regex with str.match:
pattern = '|'.join(ENSEMBL_IDs)
m = genes_df[0].str.match(pattern)
out = genes_df.loc[m, 1].tolist()
Output: ['RTN4R', 'CLTCL1', 'DGCR2']
I have a list named ChargeType and a list named TaskCode. ChargeType contains "Professional Fees", "Disbursements", and "Subcontractors". TaskCode contains say 0527014.05.01, 0527014.05.02, 0527014.07.01, 0527014.04.01, 0527014.02.01 etc. I'm trying to create a unique list from another named range (JobClass). However, I can't seem to get the wildcard to work for me.
My formula right now is {=IFERROR(INDEX(JobClass,MATCH(0,IF(ChargeType="Professional Fees",IF(TaskCode="0527014.05"&"*",COUNTIF($D$6:D6,JobClass),""),""),0)),"")} (dragged down of course). So I want it to pick out the unique items from JobClass if the ChargeType is "Professional Fees" and the TaskCode is any of the 0527014.05 items. However, it is returning blanks.
If I don't use the wildcard and instead put the entire 0527014.05.01 it works, but only gives me Intermediate I, Principal I, and Intermediate II*. But I need it to work for all of the .05's. What am I doing wrong?
PTH = TaskCode, JobClass/Ref = JobClass, BudgetType = ChargeType
Desired output in this case would be a unique list of Intermediate I, Principal I, Intermediate II, Senior I
I'm making a game of hangman. I use a list to keep track of the word that you are guessing for, and a list of blanks that you fill in. But I can't figure out what to do if for example someone's word was apple, and I guessed p.
My immediate thought was to just find if a letter is in the word twice, then figure out where it is, and when they guess that letter put it in both the first and second spot where that letter is. But I can't find
How to test if two STRINGS are duplicates in a list, and
If I were to use list.index to test where the duplicate letters are how to I find both positions instead of just one.
Create a string for your word
Create a string for user input
Cut your string into letters and keep it on a list/array
Get input
Cut input into letters and keep it on another array
Create a string = "--------" as displayed message
Using a for loop check every position in both array lists and compare them
If yourArray[i] == inputArray[i]
Then change displayedString[i] = inputArray[i] and display message then get another input
If it doesnt match leave "-" sings
Displayed the "---a--b" string
One way to do it would be to go through the list one by one and check if something comes up twice.
def isDuplicate(myList):
a = []
index = 0
for item in myList:
if type(item) == str:
if item in a:
return index
else:
a.append(item)
index += 1
return False
This function goes through the list and adds what it has seen so far into another list. Each time it also checks if the item it is looking at is already in that list, meaning it has already been seen before. If it gets through the whole list without any duplicates, it returns False.
It also keeps track of the index it is on, so it can return that index if it does find a duplicate.
Alternately, If you want to find multiple occurrences of a given string, you would use the same structure with some modifications.
def isDuplicate(myList, query):
index = 0
foundIndexes = []
for item in myList:
if item == query:
foundIndexes.append(index)
index += 1
return foundIndexes
This would return a list of the indexes of all instances of query in myList.
I'm trying to use get() to access a list element in R, but am getting an error.
example.list <- list()
example.list$attribute <- c("test")
get("example.list") # Works just fine
get("example.list$attribute") # breaks
## Error in get("example.list$attribute") :
## object 'example.list$attribute' not found
Any tips? I am looping over a vector of strings which identify the list names, and this would be really useful.
Here's the incantation that you are probably looking for:
get("attribute", example.list)
# [1] "test"
Or perhaps, for your situation, this:
get("attribute", eval(as.symbol("example.list")))
# [1] "test"
# Applied to your situation, as I understand it...
example.list2 <- example.list
listNames <- c("example.list", "example.list2")
sapply(listNames, function(X) get("attribute", eval(as.symbol(X))))
# example.list example.list2
# "test" "test"
Why not simply:
example.list <- list(attribute="test")
listName <- "example.list"
get(listName)$attribute
# or, if both the list name and the element name are given as arguments:
elementName <- "attribute"
get(listName)[[elementName]]
If your strings contain more than just object names, e.g. operators like here, you can evaluate them as expressions as follows:
> string <- "example.list$attribute"
> eval(parse(text = string))
[1] "test"
If your strings are all of the type "object$attribute", you could also parse them into object/attribute, so you can still get the object, then extract the attribute with [[:
> parsed <- unlist(strsplit(string, "\\$"))
> get(parsed[1])[[parsed[2]]]
[1] "test"
flodel's answer worked for my application, so I'm gonna post what I built on it, even though this is pretty uninspired. You can access each list element with a for loop, like so:
#============== List with five elements of non-uniform length ================#
example.list=
list(letters[1:5], letters[6:10], letters[11:15], letters[16:20], letters[21:26])
#===============================================================================#
#====== for loop that names and concatenates each consecutive element ========#
derp=c(); for(i in 1:length(example.list))
{derp=append(derp,eval(parse(text=example.list[i])))}
derp #Not a particularly useful application here, but it proves the point.
I'm using code like this for a function that calls certain sets of columns from a data frame by the column names. The user enters a list with elements that each represent different sets of column names (each set is a group of items belonging to one measure), and the big data frame containing all those columns. The for loop applies each consecutive list element as the set of column names for an internal function* applied only to the currently named set of columns of the big data frame. It then populates one column per loop of a matrix with the output for the subset of the big data frame that corresponds to the names in the element of the list corresponding to that loop's number. After the for loop, the function ends by outputting that matrix it produced.
Not sure if you're looking to do something similar with your list elements, but I'm happy I picked up this trick. Thanks to everyone for the ideas!
"Second example" / tangential info regarding application in graded response model factor scoring:
Here's the function I described above, just in case anyone wants to calculate graded response model factor scores* in large batches...Each column of the output matrix corresponds to an element of the list (i.e., a latent trait with ordinal indicator items specified by column name in the list element), and the rows correspond to the rows of the data frame used as input. Each row should presumably contain mutually dependent observations, as from a given individual, to whom the factor scores in the same row of the ouput matrix belong. Also, I feel I should add that if all the items in a given list element use the exact same Likert scale rating options, the graded response model may be less appropriate for factor scoring than a rating scale model (cf. http://www.rasch.org/rmt/rmt143k.htm).
'grmscores'=function(ColumnNameList,DataFrame) {require(ltm) #(Rizopoulos,2006)
x = matrix ( NA , nrow = nrow ( DataFrame ), ncol = length ( ColumnNameList ))
for(i in 1:length(ColumnNameList)) #flodel's magic featured below!#
{x[,i]=factor.scores(grm(DataFrame[, eval(parse(text= ColumnNameList[i]))]),
resp.patterns=DataFrame[,eval(parse(text= ColumnNameList[i]))])$score.dat$z1}; x}
Reference
*Rizopoulos, D. (2006). ltm: An R package for latent variable modelling and item response theory analyses, Journal of Statistical Software, 17(5), 1-25. URL: http://www.jstatsoft.org/v17/i05/