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QUESTION
In rows, 11:13, and in 14:16, it can be observed that there are duplicate entries in column 'C2_xsampa' for 'm:' and 'n:'. Each value in 'C2_xsampa' has two levels, Singleton or Geminate but it is not the case among 'm:' and 'n:'. This yields wrong mean values for numeric columns.
My question is: How do I filter which row is being duplicated? I have manually checked the parent dataset through which means values are obtained. All looks fine there.
Earlier, I was using subset () to rectify the 'real' errors in entry.
Data:
...ANSWER
Answered 2021-Jun-06 at 07:54You could check that the values for the two columns are unique throughout the dataset
df = df.drop_duplicates(subset=['C2_xsampa','Consonant'])
You can get the inverse df[~df]
to get the rows that are incorrect
edit just saw the r language tag
I believe distinct(select(df, C2_xsampa, Consonant))
will do
QUESTION
I have some data here for which I wish to make a scatterplot.
...ANSWER
Answered 2021-Mar-15 at 14:32Below is a potential plotting solution using {ggplot2}. If you wanted to map the color or transparency of each point to some other variable such as V2.dn
similar to the image you shared, you could for example add alpha = V2.dn
inside the aes()
of geom_point()
.
QUESTION
I wish to replace the present value in column 'Manner' for all the words containing, 'kala', 'kalla' in column 'Filename' (the first two initials denote the Speaker) from "Sonorant' to "Liquid".
My idea is that it could be performed through 'dplyr' (probably 'grep') but don't know how to. This can be done in MS Excel but it will take a great amount of time by doing it manually.
Any help will be great. Thanks in advance.
Some of my data here:
...ANSWER
Answered 2021-Mar-10 at 07:29You can check for pattern in Filename
column and turn the Manner
column to 'Liquid'
.
QUESTION
I wish to extract all columns for rows 4, 11 and so on. If you look at my posted data, my wish is to extract row values that are present immediately before an 'A' in column 'xsampa'. For example, all the columns for row 4 (that occur before row 5 that contains an 'A' in column 'xsampa'). I can manually extract them but anything better will definitely save me some labour.
Many thanks if you help me out.
...ANSWER
Answered 2021-Feb-21 at 19:10As @Jon Spring replied in the comments, the answer to this question is to use dplyr:: lead() function instead of lag(). This way, all the rows in the column 'xsampa' that contained the value 'A' will be filtered and produce the desired output.
The lag() function will simply produce rows one behind the input.
ANSWER:
QUESTION
I am trying to perform an LMER test on a dataset (original data attached), the number of rows for all columns is the same (153). However, it gives me an error when I try to fit the formula
...Error: number of levels of each grouping factor must be < number of observations (problems: Filename)
ANSWER
Answered 2020-Oct-25 at 11:11Assuming you want the speaker to be a random effect, and further assuming that the filenames are actually labelled according to the speaker's initials and the spoken phoneme, then you need to use the initials only in the first column for your random effect. Otherwise you have only a single observation at each level of your random effect, which doesn't make much sense.
Therefore if you do:
QUESTION
I would like to have a regex expression that (in java) will replace every repeated consonant into single letter, all repeated consonants but an initial "inn". I explain myself better with some examples:
asso > aso
assso > aso
assocco > asoco
innasso > innaso
I found a way to replace all repeated letters with
Pattern.compile("([^aeiou])+\1").matcher(text).replaceAll("$1")
I found a way to recognize if a word does not start with "inn":
Pattern.compile("^(?!inn).+").matcher(text).matches()
but I don't know how to merge them, ie, degeminate all geminates consonants but the initial 'nn' if the word starts with 'inn'.
Anyone can help me? (I would like to solve this with a regex, in order to apply replaceAll
)
Thank you
...ANSWER
Answered 2020-Jan-15 at 11:59I'm not sure why you must do this all with a single regexp, but if you must... try using negative lookbehind:
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