I have a variable actor which is a string and contains values like "military forces of guinea-bissau (1989-1992)" and a large range of other different values that are fairly complex. I have been using grep() to find character patterns that match different types of actors. For example I would like to code a new variable actor_type as 1 when actor contains "military forces of", doesn't contain "mutiny of", and the string variable country is also contained in the variable actor.
I am at a loss as to how to conditionally create this new variable without resorting to some type of horrible for loop. Help me!
Data looks roughly like this:
| | actor | country |
|---+----------------------------------------------------+-----------------|
| 1 | "military forces of guinea-bissau" | "guinea-bissau" |
| 2 | "mutiny of military forces of guinea-bissau" | "guinea-bissau" |
| 3 | "unidentified armed group (guinea-bissau)" | "guinea-bissau" |
| 4 | "mfdc: movement of democratic forces of casamance" | "guinea-bissau" |
if your data is in a data.frame df:
> ifelse(!grepl('mutiny of' , df$actor) & grepl('military forces of',df$actor) & apply(df,1,function(x) grepl(x[2],x[1])),1,0)
[1] 1 0 0 0
grepl returns a logical vector and this can be assigned to whatever, e.g. df$actor_type.
breaking that appart:
!grepl('mutiny of', df$actor) and grepl('military forces of', df$actor) satisfy your first two requirements. the last piece, apply(df,1,function(x) grepl(x[2],x[1])) goes row by row and greps for country in actor.
Related
I have two dataframes. One is a set of ≈4000 entries that looks similar to this:
| grade_col1 | grade_col2 |
| --- | --- |
| A-| A-|
| B | 86|
| C+| C+|
| B-| D |
| A | A |
| C-| 72|
| F | 96|
| B+| B+|
| B | B |
| A-| A-|
The other is a set of ≈700 entries that look similar to this:
| grade | scale |
| --- | --- |
| A+|100|
| A+| 99|
| A+| 98|
| A+| 97|
| A | 96|
| A | 95|
| A | 94|
| A | 93|
| A-| 92|
| A-| 91|
| A-| 90|
| B+| 89|
| B+| 88|
...and so on.
What I'm trying to do is create a new column that shows whether grade_col2 matches grade_col1 with a binary, 0-1 output (0 = no match, 1 = match). Most of grade_col2 is shown by letter grade. But every once in awhile an entry in grade_col2 was accidentally entered as a numeric grade instead. I want this match column to give me a "1" even when grade_col2 is a numeric grade instead of a letter grade. In other words, if grade_col1 is B and grade_col2 is 86, I want this to still be read as a match. Only when grade_col1 is F and grade_col2 is 96 would this not be a match (similar to when grade_col1 is B- and grade_col2 is D = not a match).
The second data frame gives me the information I need to translate between one and the other (entries between 97-100 are A+, between 93-96 are A, and so on). I just don't know how to run a script that uses this information to find matches through all ≈4000 entries. Theoretically, I could do this manually, but the real dataset is so lengthy that this isn't realistic.
I had been thinking of using nested if_else statements with dplyr. But once I got past the first "if" statement, I got stuck. I'd appreciate any help with this people can offer.
You can do this using a join.
Let your first dataframe be grades_df and your second dataframe be lookup_df, then you want something like the following:
output = grades_df %>%
# join on look up, keeping everything grades table
left_join(lookup_df, by = c(grade_col2 = "scale")) %>%
# combine grade_col2 from grades_df and grade from lookup_df
mutate(grade_col2b = ifelse(is.na(grade), grade_col2, grade)) %>%
# indicator column
mutate(indicator = ifelse(grade_col1 == grade_col2b, 1, 0))
I merged two data sets using Stata and now I need to find the fraction and number of projects matched. To do this, I am assuming that I will need to calculate two counts.
How do I get both of the counts to display at the same time, and then divide one by the other?
Below is an example of my _merge variable:
4022. | master only (1) |
4023. | matched (3) |
4024. | using only (2) |
4025. | using only (2) |
4026. | using only (2) |
4027. | matched (3) |
4028. | matched (3) |
4029. | matched (3) |
4030. | matched (3) |
I would first like to count and store all of the variables under _merge, and then count those that don't say "master only". Then divide the two by each other.
For example:
count1 count2 fraction
6019 4020 .66 (4020/6019)
With count1 being everything under _merge, while count2 being everything that was matched (excludes master only).
Using the following toy example:
clear
webuse autosize
merge 1:1 make using http://www.stata-press.com/data/r14/autoexpense
First it is a good idea to confirm the value which corresponds to "master only":
list _merge
+-----------------+
| _merge |
|-----------------|
1. | matched (3) |
2. | matched (3) |
3. | matched (3) |
4. | master only (1) |
5. | matched (3) |
|-----------------|
6. | matched (3) |
+-----------------+
list _merge, nolabel
+--------+
| _merge |
|--------|
1. | 3 |
2. | 3 |
3. | 3 |
4. | 1 |
5. | 3 |
|--------|
6. | 3 |
+--------+
Then generate the three variables by first counting the relevant observations and dividing:
count if _merge
generate count1 = r(N)
count if _merge != 1
generate count2 = r(N)
generate fraction = count2 / count1
display count1
6
display count2
5
display fraction
1.2
I know this is a simple question so I apologize in advance.
If I have a dataframe like this:
| name | count | class |
|-------|-------|-------------|
| bob | 1 | first grade |
| adam | 5 | college |
| suzie | 7 | high school |
and I want to reorder the rows by class, as in:
| name | count | class |
|-------|-------|-------------|
| bob | 1 | first grade |
| suzie | 7 | high school |
| adam | 5 | college |
I can't use order() since I don't want the class reordered alphabetically.
I tried this, but it failed:
target <- c("first grade", "high school", "college")
df[match(target, df$class),]
This should be straightforward...but reordering is usually reserved for when the values in the columns have some sort of alphanumeric structure. Here, the structure is to be defined by me.
I suppose I could append a new column, with number assignments for class, then sort by that. But there has got to be a more graceful way??
Make class a factor with the levels in the order you want, then use order().
df$class = factor(df$class, levels = target)
df[order(df$class), ]
I think you can do this via an ordered factor.
First create a factor variable from your variable of interest
d <- df$class
Then order the factor by the order you wish
x <- ordered(factor(d), levels=c('first grade','high school','college'))
Then use this to order your df
df[order(x),]
Job done, go play a board game.
Your match needs to be modified a little to work:
df[order(match(df$class, target)),]
I have a spreadsheet where dates are being recorded in regards to individuals, with additional data, as such:
Tom | xyz | 5/2/2012
Dick | foo | 5/2/2012
Tom | bar | 6/1/2012
On another sheet there is a line in which I want to be able to put in the name, such as Tom, and retrieve on the following cell through a formula the data for the LAST (most recent by date) entry in the first sheet. So the first sheet is a log, and the second sheet displays the most recent one. In the following example, the first cell is entered and the remaining are formulas displaying data from the first sheet:
Tom | bar | 6/1/2012
and so on, showing the latest dated entry in the log.
I'm stumped, any ideas?
If you only need to do a single lookup, you can do that by adding two new columns in your log sheet:
Sheet1
| A | B | C | D | E | F
1 | Tom | xyz | 6/2/2012 | | * | *
2 | Dick | foo | 5/2/2012 | | * | *
3 | Tom | bar | 6/1/2012 | | * | *
Sheet2
| A | B | C
1 | Tom | =Sheet1.E1 | =Sheet1.F1
*(E1) = =IF(AND($A1=Sheet2.$A$1;E2=0);B1;E2)
(i.e. paste the formula above in E1, then copy/paste it in the other cells with *)
Explanation: if A is not what you're looking for, go for the next; if it is, but there is a non-empty next, go for the next; otherwise, get it. This way you're selecting the last one corresponding to your search. I'm assuming you want the last entry, not "the one with the most recent date", since that's what you asked in your example. If I interpreted your question wrong, please update it and I can try to provide a better answer.
Update: If the log dates can be out of order, here's how you get the last entry:
*(F1) = =IF(AND($A1=Sheet2.$A$1;C1>=F2);C1;F2)
*(E1) = =IF(C1=F1;B1;E2)
Here I just replaced the test F2=0 (select next if non-empty) for C1>=F2 (select next if more recent) and, for the other column, select next if the first test also did so.
Disclaimer: I'm very inexperienced with spreadsheets, the solution above is ugly but gets the job done. For instance, if you wanted a 2nd row in Sheet2 to do another lookup, you'd need to add two more columns to Sheet1, etc.
Months ago I ended up with a sub statement that originally worked with my input data. It has since stopped working causing me to re-examine my ugly process. I hate to share it but it accomplished several things at once:
active$id[grep("CIR",active$description)] <- sub(".*CIR0*(\\d+).*","\\1",active$description[grep("CIR",active$description)],perl=TRUE)
This statement created a new id column by finding rows that had an id embedded in the description column. The sub statement would find the number following a "CIR0" and populate the id column iff there was an id within a row's description. I recognize it is inefficient with the embedded grep subsetting either side of the assignment.
Is there a way to have a 'sub' replacement be NA or empty if the pattern does not match? I feel like I'm missing something very simple but ask for the community's assistance. Thank you.
Example with the results of creating an id column:
| name | id | description |
|------+-----+-------------------|
| a | 343 | Here is CIR00343 |
| b | | Didn't have it |
| c | 123 | What is CIR0123 |
| d | | CIR lacks a digit |
| e | 452 | CIR452 is next |
I was struggling with the same issue a few weeks ago. I ended up using the str_match function from the stringr package. It returns NA if the target string is not found. Just make sure you subset the result correctly. An example:
library(stringr)
str = "Little_Red_Riding_Hood"
sub(".*(Little).*","\\1",str) # Returns 'Little'
sub(".*(Big).*","\\1",str) # Returns 'Little_Red_Riding_Hood'
str_match(str,".*(Little).*")[1,2] #Returns 'Little'
str_match(str,".*(Big).*")[1,2] # Returns NA
I think in this case you could try using ifelse(), i.e.,
active$id[grep("CIR",active$description)] <- ifelse(match, replacement, "")
where match should evaluate to true if there's a match, and replacement is what that element would be replaced with in that case. Likewise, if match evaluates to false, that element's replaced with an empty string (or NA if you prefer).