Converting a data set into a data frame

Question

I have this data set

df<-
structure(list(x = list(structure(c(1.2956002895122, 0.84856971403248, 
0.318587376385379, 1.52311609250527, 0.761866777256873, 1.34942921531382, 
1.32182157459345, 0.164293091865351, 1.33391849692185, 1.1741444035705
), dim = c(5L, 2L), dimnames = list(NULL, c("Consumption", "Income"
))), structure(c(0.84804804783009, 0.454659301099081, -0.785771762432506, 
0.746346341726964, 1.07546066123554, 2.36782435778472, 0.33145722604832, 
-1.16704208085918, -0.505189833907774, -0.0305510764003426), dim = c(5L, 
2L), dimnames = list(NULL, c("Consumption", "Income"))), structure(c(1.67206824624292, 
0.621301929142792, -2.64748116335658, 1.06382925864556, 0.481820198662883, 
0.977532978223718, 0.12858898496739, -0.929583796669093, 1.03055091162915, 
1.75054296893639), dim = c(5L, 2L), dimnames = list(NULL, c("Consumption", 
"Income"))), structure(c(1.06124884050636, 1.5695107316969, -0.71012903796485, 
0.697670432455871, 0.441910553257345, 1.131314331505, 1.58797564122333, 
-0.0245687154568635, 1.26341241328479, -0.701603888772303), dim = c(5L, 
2L), dimnames = list(NULL, c("Consumption", "Income"))), structure(c(-0.320387157182086, 
0.909612038039385, -0.30534801455308, 0.13028454605271, 1.04548318778476, 
0.744569146062672, 0.157712771975432, -0.47140382117885, -0.757295165153137, 
1.30319365700164), dim = c(5L, 2L), dimnames = list(NULL, c("Consumption", 
"Income"))), structure(c(1.1328615935615, 1.18848732447009, -1.19259137763868, 
1.61383333543976, 0.112684975627695, -0.294284500868068, 0.653900926135023, 
0.0179116410157527, 2.3485206187256, 0.561897344846424), dim = c(5L, 
2L), dimnames = list(NULL, c("Consumption", "Income"))), structure(c(0.97523495340453, 
2.13698353093441, 0.766663720261808, 0.693029332670238, 0.704645561153127, 
1.29154523355534, 4.14154953485982, -0.454659021170493, 0.55475024788991, 
0.63410937974327), dim = c(5L, 2L), dimnames = list(NULL, c("Consumption", 
"Income"))))), row.names = c(NA, -7L), class = "data.frame")

How can I put this to a data frame? Right now each row is considered as 1 value and I can not access the elements of each row.

M-- · Accepted Answer · 2026-04-03 17:55:33Z

Your df has list columns where each row contains a matrix. So you need to unpack them. Here's an approach that gives you a tidy dataframe:

do.call(rbind, lapply(seq_along(df$x), function(i) {
  data.frame(group = i, df$x[[i]])
}))

#>    group Consumption      Income
#> 1      1   1.2956003  1.34942922
#> 2      1   0.8485697  1.32182157
#> 3      1   0.3185874  0.16429309
#> 4      1   1.5231161  1.33391850
#> 5      1   0.7618668  1.17414440
#> ...
#> 30     6   0.1126850  0.56189734
#> 31     7   0.9752350  1.29154523
#> 32     7   2.1369835  4.14154953
#> 33     7   0.7666637 -0.45465902
#> 34     7   0.6930293  0.55475025
#> 35     7   0.7046456  0.63410938

^{Created on 2026-04-03 with reprex v2.1.1}

or using {dplyr}:

library(dplyr)

df |>
  mutate(group = row_number()) |>
  reframe(
    group = group,
    as.data.frame(x),
    .by = group
  )

If you want it in wide format:

library(tidyverse)

df |>
  mutate(group = row_number(), .before = 1,
         x = map(x, ~set_names(c(.x), 
                               paste0(rep(colnames(.x), each = nrow(.x)), "_", 
                                      rep(seq_len(nrow(.x)), ncol(.x)))))) |>
  unnest_wider(x)
#> # A tibble: 7 × 11
#>   group Consumption_1 Consumption_2 Consumption_3 Consumption_4 Consumption_5
#>   <int>         <dbl>         <dbl>         <dbl>         <dbl>         <dbl>
#> 1     1         1.30          0.849         0.319         1.52          0.762
#> 2     2         0.848         0.455        -0.786         0.746         1.08 
#> 3     3         1.67          0.621        -2.65          1.06          0.482
#> 4     4         1.06          1.57         -0.710         0.698         0.442
#> 5     5        -0.320         0.910        -0.305         0.130         1.05 
#> 6     6         1.13          1.19         -1.19          1.61          0.113
#> 7     7         0.975         2.14          0.767         0.693         0.705
#> # ℹ 5 more variables: Income_1 <dbl>, Income_2 <dbl>, Income_3 <dbl>,
#> #   Income_4 <dbl>, Income_5 <dbl>

^{Created on 2026-04-03 with reprex v2.1.1}

Or in {base}:

data.frame(
  group = seq_len(nrow(df)),
  do.call(rbind, lapply(df$x, function(m) {
    setNames(
      c(m), 
      paste0(rep(colnames(m), each = nrow(m)), "_", 
             rep(seq_len(nrow(m)), ncol(m)))
    )
  }))
)

Friede · Accepted Answer · 2026-04-07 19:45:10Z

With {collapse}

collapse::unlist2d(l=df$x)

e.g.

collapse::unlist2d(l=df$x) |>  
  head()

  .id Consumption    Income
1   1   1.2956003 1.3494292
2   1   0.8485697 1.3218216
3   1   0.3185874 0.1642931
4   1   1.5231161 1.3339185
5   1   0.7618668 1.1741444
6   2   0.8480480 2.3678244

With idcols-argument we can specify the name (defaults to .id). {collapse} is generally considered a rapidfly fast package, it also ships collapse::pivot-function for reshaping.

With {base} (from R Core)

If the group column (.id) is not needed, simply

df$x |> 
  do.call(what='rbind') |> 
  as.data.frame.matrix()

If needed, I tend to use a Map + [<- construct like

df$x |>
  lapply(data.frame) |> 
  ( \(.) Map(`[<-`, ., '.id', value=seq_along(df$x)) )() |> 
  do.call(what='rbind')

For wide format

data.frame(df$x)
## calls `as.data.frame()` which has `.list`-method

  Consumption    Income Consumption.1    Income.1 Consumption.2   Income.2 Consumption.3    Income.3 Consumption.4   Income.4 Consumption.5    Income.5 Consumption.6   Income.6
1   1.2956003 1.3494292     0.8480480  2.36782436     1.6720682  0.9775330     1.0612488  1.13131433    -0.3203872  0.7445691      1.132862 -0.29428450     0.9752350  1.2915452
2   0.8485697 1.3218216     0.4546593  0.33145723     0.6213019  0.1285890     1.5695107  1.58797564     0.9096120  0.1577128      1.188487  0.65390093     2.1369835  4.1415495
3   0.3185874 0.1642931    -0.7857718 -1.16704208    -2.6474812 -0.9295838    -0.7101290 -0.02456872    -0.3053480 -0.4714038     -1.192591  0.01791164     0.7666637 -0.4546590
4   1.5231161 1.3339185     0.7463463 -0.50518983     1.0638293  1.0305509     0.6976704  1.26341241     0.1302845 -0.7572952      1.613833  2.34852062     0.6930293  0.5547502
5   0.7618668 1.1741444     1.0754607 -0.03055108     0.4818202  1.7505430     0.4419106 -0.70160389     1.0454832  1.3031937      0.112685  0.56189734     0.7046456  0.6341094

might be enough.

Notes

df <- <sth> masks stats::df-function which ships with R Core. From the head of the corresponding help file

FDist {stats} R Documentation The F Distribution Description Density, distribution function, quantile function and random generation for the F distribution with df1 and df2 degrees of freedom (and optional non-centrality parameter ncp). Usage df(x, df1, df2, ncp, log = FALSE) ...

Therefore, I recommend to assign to either metasyntactic variable names, e.g. foobar, foo, bar, baz, quux, xyzzy, ...,--personally I like .data, x, X those are rather R-specific--or something data-specific, e.g. cshd for cross-sectional household data.

Where does the data come from? It’s possible there’s an issue with the import process, and we may currently be dealing with an XY-problem.
Some tidy data references would encourage us to keep the data in long format. A well-known source.

Andre Wildberg · Accepted Answer · 2026-04-07 14:25:08Z

Using unlist with as.data.frame

as.data.frame(unlist(df, recursive = FALSE))

output:

  x1.Consumption x1.Income x2.Consumption   x2.Income x3.Consumption  x3.Income
1      1.2956003 1.3494292      0.8480480  2.36782436      1.6720682  0.9775330
2      0.8485697 1.3218216      0.4546593  0.33145723      0.6213019  0.1285890
3      0.3185874 0.1642931     -0.7857718 -1.16704208     -2.6474812 -0.9295838
4      1.5231161 1.3339185      0.7463463 -0.50518983      1.0638293  1.0305509
5      0.7618668 1.1741444      1.0754607 -0.03055108      0.4818202  1.7505430
  x4.Consumption   x4.Income x5.Consumption  x5.Income x6.Consumption
1      1.0612488  1.13131433     -0.3203872  0.7445691       1.132862
2      1.5695107  1.58797564      0.9096120  0.1577128       1.188487
3     -0.7101290 -0.02456872     -0.3053480 -0.4714038      -1.192591
4      0.6976704  1.26341241      0.1302845 -0.7572952       1.613833
5      0.4419106 -0.70160389      1.0454832  1.3031937       0.112685
    x6.Income x7.Consumption  x7.Income
1 -0.29428450      0.9752350  1.2915452
2  0.65390093      2.1369835  4.1415495
3  0.01791164      0.7666637 -0.4546590
4  2.34852062      0.6930293  0.5547502
5  0.56189734      0.7046456  0.6341094

ThomasIsCoding · Accepted Answer · 2026-04-07 18:59:42Z

With base R you can use basic operations like cbind + rep + nrow + rbind + Reduce

with(df,
     cbind(
      data.frame(grp = rep(seq_along(x), lapply(x,nrow))),
      Reduce(rbind,x)))

which gives

   grp Consumption      Income
1    1   1.2956003  1.34942922
2    1   0.8485697  1.32182157
3    1   0.3185874  0.16429309
4    1   1.5231161  1.33391850
5    1   0.7618668  1.17414440
6    2   0.8480480  2.36782436
7    2   0.4546593  0.33145723
8    2  -0.7857718 -1.16704208
9    2   0.7463463 -0.50518983
10   2   1.0754607 -0.03055108
11   3   1.6720682  0.97753298
12   3   0.6213019  0.12858898
13   3  -2.6474812 -0.92958380
14   3   1.0638293  1.03055091
15   3   0.4818202  1.75054297
16   4   1.0612488  1.13131433
17   4   1.5695107  1.58797564
18   4  -0.7101290 -0.02456872
19   4   0.6976704  1.26341241
20   4   0.4419106 -0.70160389
21   5  -0.3203872  0.74456915
22   5   0.9096120  0.15771277
23   5  -0.3053480 -0.47140382
24   5   0.1302845 -0.75729517
25   5   1.0454832  1.30319366
26   6   1.1328616 -0.29428450
27   6   1.1884873  0.65390093
28   6  -1.1925914  0.01791164
29   6   1.6138333  2.34852062
30   6   0.1126850  0.56189734
31   7   0.9752350  1.29154523
32   7   2.1369835  4.14154953
33   7   0.7666637 -0.45465902
34   7   0.6930293  0.55475025
35   7   0.7046456  0.63410938

Another interesting way of presenting the dataframe format might be

do.call(Reduce, c(cbind,df))

which gives

     Consumption    Income Consumption      Income Consumption     Income
[1,]   1.2956003 1.3494292   0.8480480  2.36782436   1.6720682  0.9775330
[2,]   0.8485697 1.3218216   0.4546593  0.33145723   0.6213019  0.1285890
[3,]   0.3185874 0.1642931  -0.7857718 -1.16704208  -2.6474812 -0.9295838
[4,]   1.5231161 1.3339185   0.7463463 -0.50518983   1.0638293  1.0305509
[5,]   0.7618668 1.1741444   1.0754607 -0.03055108   0.4818202  1.7505430
     Consumption      Income Consumption     Income Consumption      Income
[1,]   1.0612488  1.13131433  -0.3203872  0.7445691    1.132862 -0.29428450
[2,]   1.5695107  1.58797564   0.9096120  0.1577128    1.188487  0.65390093
[3,]  -0.7101290 -0.02456872  -0.3053480 -0.4714038   -1.192591  0.01791164
[4,]   0.6976704  1.26341241   0.1302845 -0.7572952    1.613833  2.34852062
[5,]   0.4419106 -0.70160389   1.0454832  1.3031937    0.112685  0.56189734
     Consumption     Income
[1,]   0.9752350  1.2915452
[2,]   2.1369835  4.1415495
[3,]   0.7666637 -0.4546590
[4,]   0.6930293  0.5547502
[5,]   0.7046456  0.6341094

M-- · Accepted Answer · 2026-04-03 17:17:06Z

3

If you use the dplyr package, there is an even shorter and more readable way to achieve the exact same result:

library(dplyr)

# Convert matrices to data frames and bind them together, 

# automatically creating an ID column for the groups

clean_df <- bind_rows(lapply(df$x, as.data.frame), .id = "group")

edited Apr 3 at 17:17

M--

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answered Apr 3 at 17:11

Vasiliy Enyutin

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