Since you already know the formula, it should be easy enough to create a function to do the calculation for you.

Here, I've created a basic function to get you started. The function takes four arguments:

• frequencies: A vector of frequencies ("number" in your first example)
• intervals: A 2-row matrix with the same number of columns as the length of frequencies, with the first row being the lower class boundary, and the second row being the upper class boundary. Alternatively, "intervals" may be a column in your data.frame, and you may specify sep (and possibly, trim) to have the function automatically create the required matrix for you.
• sep: The separator character in your "intervals" column in your data.frame.
• trim: A regular expression of characters that need to be removed before trying to coerce to a numeric matrix. One pattern is built into the function: trim = "cut". This sets the regular expression pattern to remove (, ), [, and ] from the input.

Here's the function (with comments showing how I used your instructions to put it together):

GroupedMedian <- function(frequencies, intervals, sep = NULL, trim = NULL) {
  # If "sep" is specified, the function will try to create the 
  #   required "intervals" matrix. "trim" removes any unwanted 
  #   characters before attempting to convert the ranges to numeric.
  if (!is.null(sep)) {
    if (is.null(trim)) pattern <- ""
    else if (trim == "cut") pattern <- "\\[|\\]|\\(|\\)"
    else pattern <- trim
    intervals <- sapply(strsplit(gsub(pattern, "", intervals), sep), as.numeric)
  }

  Midpoints <- rowMeans(intervals)
  cf <- cumsum(frequencies)
  Midrow <- findInterval(max(cf)/2, cf) + 1
  L <- intervals[1, Midrow]      # lower class boundary of median class
  h <- diff(intervals[, Midrow]) # size of median class
  f <- frequencies[Midrow]       # frequency of median class
  cf2 <- cf[Midrow - 1]          # cumulative frequency class before median class
  n_2 <- max(cf)/2               # total observations divided by 2

  unname(L + (n_2 - cf2)/f * h)
}

Here's a sample data.frame to work with:

mydf <- structure(list(salary = c("1500-1600", "1600-1700", "1700-1800", 
    "1800-1900", "1900-2000", "2000-2100", "2100-2200", "2200-2300", 
    "2300-2400", "2400-2500"), number = c(110L, 180L, 320L, 460L, 
    850L, 250L, 130L, 70L, 20L, 10L)), .Names = c("salary", "number"), 
    class = "data.frame", row.names = c(NA, -10L))
mydf
#       salary number
# 1  1500-1600    110
# 2  1600-1700    180
# 3  1700-1800    320
# 4  1800-1900    460
# 5  1900-2000    850
# 6  2000-2100    250
# 7  2100-2200    130
# 8  2200-2300     70
# 9  2300-2400     20
# 10 2400-2500     10

Now, we can simply do:

GroupedMedian(mydf$number, mydf$salary, sep = "-")
# [1] 1915.294

Here's an example of the function in action on some made up data:

set.seed(1)
x <- sample(100, 100, replace = TRUE)
y <- data.frame(table(cut(x, 10)))
y
#           Var1 Freq
# 1   (1.9,11.7]    8
# 2  (11.7,21.5]    8
# 3  (21.5,31.4]    8
# 4  (31.4,41.2]   15
# 5    (41.2,51]   13
# 6    (51,60.8]    5
# 7  (60.8,70.6]   11
# 8  (70.6,80.5]   15
# 9  (80.5,90.3]   11
# 10  (90.3,100]    6

### Here's GroupedMedian's output on the grouped data.frame...
GroupedMedian(y$Freq, y$Var1, sep = ",", trim = "cut")
# [1] 49.49231

### ... and the output of median on the original vector
median(x)
# [1] 49.5

By the way, with the sample data that you provided, where I think there was a mistake in one of your ranges (all were separated by dashes except one, which was separated by a comma), since strsplit uses a regular expression by default to split on, you can use the function like this:

x<-c(110,180,320,460,850,250,130,70,20,10)
colnames<-c("numbers")
rownames<-c("[1500-1600]","(1600-1700]","(1700-1800]","(1800-1900]",
            "(1900-2000]"," (2000,2100]","(2100-2200]","(2200-2300]",
            "(2300-2400]","(2400-2500]")
y<-matrix(x,nrow=length(x),dimnames=list(rownames,colnames))
GroupedMedian(y[, "numbers"], rownames(y), sep="-|,", trim="cut")
# [1] 1915.294

library(dplyr)
dat%>%
group_by(custid)%>% 
summarise(Mean=mean(value), Max=max(value), Min=min(value), Median=median(value), Std=sd(value))
#  custid     Mean Max Min Median      Std
#1      1 2.666667   5   1    2.5 1.632993
#2      2 5.500000  10   1    5.5 6.363961
#3      3 2.666667   5   1    2.0 2.081666

For bigger datasets, data.table would be faster

setDT(dat)[,list(Mean=mean(value), Max=max(value), Min=min(value), Median=as.numeric(median(value)), Std=sd(value)), by=custid]
#   custid     Mean Max Min Median      Std
#1:      1 2.666667   5   1    2.5 1.632993
#2:      2 5.500000  10   1    5.5 6.363961
#3:      3 2.666667   5   1    2.0 2.081666

I find it amazing that noone has suggested aggregate yet, seeing as it is the simple, base R function included for these sorts of tasks. E.g.:

aggregate(. ~ genotype, data=dat, FUN=median)

#  genotype DIV3  DIV4
#1      HET  1.4  3.20
#2       WT 23.9 25.25

Using R:

library(tidyverse)

Data %>% group_by(classes, states) %>% summarize(Median=median(score, na.rm = TRUE/FALSE))

One solution can be achieved using dplyr and following below mentioned steps. Please find comments in code below for approach.

Note: It seems that sample data from OP is not very meaningful as such.

library(dplyr)

df %>% filter(stuff > 0) %>%  #First filter out for stuff > 0 which of our interest
  group_by(ItemRelation, num, year) %>%
    mutate(m = median(stuff[action==1]),
           m0 = median(tail(stuff[action==0], 5))) %>%  # Calculate m and m0 for all rows
  filter(action == 1) %>%  # Now keep only rows with action == 1
  mutate(m = m-m0) %>%
  select(-Dt,-m0,-action)

# # A tibble: 4 x 5
# # Groups: ItemRelation, num, year [2]
# ItemRelation stuff   num  year     m
# <int> <int> <int> <int> <dbl>
# 1       158043   400  1459  2018  -450
# 2       158043   700  1459  2018  -450
# 3          234   400  1459  2018  -450
# 4          234   700  1459  2018  -450

Because this is essentially a duplicate, I address a few issues that are do not explicitly overlap the related question or answer:

If a class has cumulative frequency .5, then the median is at the boundary of that class and the next larger one.

If $N$ is large (really the only case where this method is generally successful), there is little difference between $N/2$ and $(N+1)/2$ in the formula. All references I checked use $N/2$.

Before computers were widely available, large datasets were customarily reduced to categories (classes) and plotted as histograms. Then the histograms were used to approximate the mean, variance, median, and other descriptive measures. Nowadays, it is best just to use a statistical computer package to find exact values of all measures.

One remaining application is to try to re-claim the descriptive measures from grouped data or from a histogram published in a journal. These are cases in which the original data are no longer available.

This procedure to approximate the sample median from grouped data $assumes$ that data are distributed in roughly a uniform fashion throughout the median interval. Then it uses interpolation to approximate the median. (By contrast, methods to approximate the sample mean and sample variance from grouped data one assumes that all obseervations are concentrated at their class midpoints.)

If I understood your question correctly you're going to want to do something like this:

# Your gestational data:
gestational_data <- data.frame(GA_weeks = c(20:26),
                               num_infants_born = c(16,22,34,45,60,67,94))

# See the apply() documentation by running 
# ?apply

apply(gestational_data,
      1,
      function(x){
        rep(x[1],x[2])
      }) |>
  unlist()|>
  median()

We can try with dplyr

library(dplyr)    
Clean1 <- Clean[rep(1:nrow(Clean), Clean$Frequency),]
Clean1 %>%
      group_by(State) %>%
      summarise(Median = median(medicare_average_payment))

Or using data.table

library(data.table)
setDT(Clean)[, .(Median = median(rep(medicare_average_payment, Frequency))) , State]