Create a Data Frame of Unequal Lengths

Create a Data Frame of Unequal Lengths

Sorry this isn't exactly what you asked, but I think there may be another way to get what you want.

First, if the vectors are different lengths, the data isn't really tabular, is it? How about just save it to different CSV files? You might also try ascii formats that allow storing multiple objects (json, XML).

If you feel the data really is tabular, you could pad on NAs:

> x = 1:5
> y = 1:12
> max.len = max(length(x), length(y))
> x = c(x, rep(NA, max.len - length(x)))
> y = c(y, rep(NA, max.len - length(y)))
> x
 [1]  1  2  3  4  5 NA NA NA NA NA NA NA
> y
 [1]  1  2  3  4  5  6  7  8  9 10 11 12

If you absolutely must make a data.frame with unequal columns you could subvert the check, at your own peril:

> x = 1:5
> y = 1:12
> df = list(x=x, y=y)
> attributes(df) = list(names = names(df),
    row.names=1:max(length(x), length(y)), class='data.frame')
> df
      x  y
1     1  1
2     2  2
3     3  3
4     4  4
5     5  5
6  <NA>  6
7  <NA>  7
 [ reached getOption("max.print") -- omitted 5 rows ]]
Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
  corrupt data frame: columns will be truncated or padded with NAs

How do you make a data frame with unequal row lengths?

This is a better version of you function that does not remove any NA from your data:

(However, the function will still trip on non numeric values for x, or in cases where scale and center are both FALSE. But one could ask oneself why a scale function needs a scale yes or no parameter??)

MyScale <- function (x, scale, center){
  meanofdata <- mean(x, na.rm = TRUE)
  stdofdata <- sd(x, na.rm = TRUE)
  
  if (scale==TRUE){
    calcvec <- (x - meanofdata)/stdofdata 
    return(calcvec)
  }else if (center ==TRUE){
    centervec <- x - meanofdata
    return(centervec)
  }
} 

Create dataframe with columns of unequal length from other dataframes

You could, also, "subset" each dataframe like df[nrow(df) + n,] in order to insert NAs:

#dataframes of different rows
long <- data.frame(accepted = rnorm(15, 2000), cost = rnorm(15,5000))
long2 <- data.frame(accepted = rnorm(10, 2000), cost = rnorm(10,5000))
long3 <- data.frame(accepted = rnorm(12, 2000), cost = rnorm(12,5000))

#insert all dataframes in list to manipulate
myls <- list(long, long2, long3)

#maximum number of rows
max.rows <- max(nrow(long), nrow(long2), nrow(long3))

#insert the needed `NA`s to each dataframe
new_myls <- lapply(myls, function(x) { x[1:max.rows,] })

#create  wanted dataframe
do.call(cbind, lapply(new_myls, `[`, "accepted"))

#   accepted accepted accepted
#1  2001.581 1999.014 2001.810
#2  2000.071 2000.033 2000.588
#3  1999.931 2000.188 2000.833
#4  1998.467 1999.891 1997.645
#5  2000.682 2000.144 1999.639
#6  1999.693 1999.341 1998.959
#7  2000.222 1998.939 2002.271
#8  1999.104 1998.530 1997.600
#9  1998.435 2001.496 2001.129
#10 1998.160 2000.729 2001.602
#11 1999.267       NA 1999.733
#12 2000.048       NA 2001.431
#13 1999.504       NA       NA
#14 2000.660       NA       NA
#15 2000.160       NA       NA

How to create a data frame using two unequal length of lists

Use np.repeat like this:

df = pd.DataFrame({'table1': column1,
                   'table2': np.repeat(column2, len(column1) // len(column2))})
print(df)

# Output:
   table1 table2
0      30    bat
1      40    bat
2      50    bat
3      60   ball
4      90   ball
5      20   ball
6      30   tent
7      20   tent
8      30   tent

Create pandas dataframe from list of lists with unequal lengths and NaN values

Just make all of your list elements to be lists:

pd.DataFrame([x if isinstance(x, list) else [x] for x in a])

To rename columns as given in the original example you could use:

pd.DataFrame([x if isinstance(x, list) else [x] for x in a]).rename(columns = lambda x: f"col_{x+1}")

Which gives:

   col_1  col_2  col_3  col_4  col_5
0    1.0    NaN    NaN    NaN    NaN
1    1.0    2.0    NaN    NaN    NaN
2    1.0    2.0    3.0    NaN    NaN
3    NaN    NaN    NaN    NaN    NaN
4    1.0    2.0    3.0    4.0    NaN
5    NaN    NaN    NaN    NaN    NaN
6    1.0    2.0    3.0    4.0    5.0

Create a pandas dataframe from a nested lists of unequal lengths

The zip_longest function from itertools does this:

>>> import itertools, pandas
>>> pandas.DataFrame((_ for _ in itertools.zip_longest(*nest)), columns=['aa', 'bb', 'cc'])
    aa    bb    cc
0  aa1   bb1   cc1
1  aa2   bb2   cc2
2  aa3   bb3   cc3
3  aa4   bb4  None
4  aa5  None  None

If you have an older version of pandas you may need to wrap zip_longest in a list constructor. On older Python you may need to call izip_longest instead of zip_longest.

Combining vectors of unequal length into a data frame

I think that you may be approaching this the wrong way:

If you have time series of unequal length then the absolute best thing to do is to keep them as time series and merge them. Most time series packages allow this. So you will end up with a multi-variate time series and each value will be properly associated with the same date.

So put your time series into zoo objects, merge them, then use my qplot.zoo function to plot them. That will deal with switching from zoo into a long data frame.

Here's an example:

> z1 <- zoo(1:8, 1:8)
> z2 <- zoo(2:8, 2:8)
> z3 <- zoo(4:8, 4:8)
> nm <- list("z1", "z2", "z3")
> z <- zoo()
> for(i in 1:length(nm)) z <- merge(z, get(nm[[i]]))
> names(z) <- unlist(nm)
> z
  z1 z2 z3
1  1 NA NA
2  2  2 NA
3  3  3 NA
4  4  4  4
5  5  5  5
6  6  6  6
7  7  7  7
8  8  8  8
> 
> x.df <- data.frame(dates=index(x), coredata(x))
> x.df <- melt(x.df, id="dates", variable="val")
> ggplot(na.omit(x.df), aes(x=dates, y=value, group=val, colour=val)) + geom_line() + opts(legend.position = "none")

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