title: I have three lists with a vector of values, how can I put all three list's values at the same index into an existing dataframe?
body: <p>A fully reproducible example. The problem is now the new columns created is in c() of all the values instead of each value being in it's individual row.</p>
  <pre><code>date = seq(as.Date(&quot;2019/01/01&quot;), by = &quot;month&quot;, length.out = 48)
  
  productB = rep(&quot;B&quot;,48)
  productB = rep(&quot;B&quot;,48)
  productA = rep(&quot;A&quot;,48)
  productA = rep(&quot;A&quot;,48)
  
  subproducts1=rep(&quot;1&quot;,48)
  subproducts2=rep(&quot;2&quot;,48)
  subproductsx=rep(&quot;x&quot;,48)
  subproductsy=rep(&quot;y&quot;,48)
  
  b1 &lt;- c(rnorm(30,5), rep(0,18))
  b2 &lt;- c(rnorm(30,5), rep(0,18))
  b3 &lt;-c(rnorm(30,5), rep(0,18))
  b4 &lt;- c(rnorm(30,5), rep(0,18))
  </code></pre>
  <p>Created the dataframe below</p>
  <pre><code>dfone &lt;- data.frame(&quot;date&quot;= rep(date,4),
                  &quot;product&quot;= c(rep(productB,2),rep(productA,2)),
                  &quot;subproduct&quot;= 
  c(subproducts1,subproducts2,subproductsx,subproductsy),
                  &quot;actuals&quot;= c(b1,b2,b3,b4))
  
  export_df &lt;- split(dfone[1:4], dfone[3])
  
  # Creation of data frames based off UNIQUE SUBPRODUCTS 
  dummy_list &lt;- split(dfone[1:4], dfone[3]) %&gt;% lapply( function(x) 
  x[(names(x) %in% c(&quot;date&quot;, &quot;actuals&quot;))])
  dummy_list &lt;-  lapply(dummy_list, function(x) { x[&quot;date&quot;] &lt;- NULL; x })
  
  
  list_dfs &lt;- list()
  for (i in 1:length(unique(dfone$subproduct))) {
    #assign(paste0(&quot;df&quot;, i), as.data.frame(dummy_list[[i]]))
    list_dfs &lt;-append(list_dfs,dummy_list[[i]])
  }
  
  combined_dfs &lt;- Reduce(function(x, y) merge(x, y, all = TRUE,  
  by='date'), list(list_dfs))
  </code></pre>
  <p>Creating the time series</p>
  <pre><code>list_ts &lt;- lapply(list_dfs, function(t) 
  ts(t,start=c(2019,1),end=c(2021,6), frequency = 12)) %&gt;%
    lapply( function(t) ts_split(t,sample.out=(0.2*length(t))))    # 
  creates my train test split
  list_ts &lt;- do.call(&quot;rbind&quot;, list_ts)  #Creates a list of time series
  </code></pre>
  <p>Creating the models</p>
  <pre><code>model_tune_ses1 &lt;- lapply(list_ts[1:(length(list_ts)/2)], function(x) 
  forecast::forecast(ses(x,h=24,alpha=0.1))) 
  model_tune_ses1 &lt;- lapply(model_tune_ses1, &quot;[&quot;,  c(&quot;mean&quot;))
  
  model_tune_ses2 &lt;- lapply(list_ts[1:(length(list_ts)/2)], function(x) 
  forecast::forecast(ses(x,h=24,alpha=0.2))) 
  model_tune_ses2 &lt;- lapply(model_tune_ses2, &quot;[&quot;,  c(&quot;mean&quot;))
  
  model_trp_holt_mult &lt;- lapply(list_ts[1:(length(list_ts)/2)], 
  function(x)
    forecast::forecast( HoltWinters(x,seasonal=&quot;multiplicative&quot;),h=24))
  
  model_trp_holt_mult &lt;- lapply(model_trp_holt_mult, &quot;[&quot;,  c(&quot;mean&quot;))
  lst1 &lt;- do.call(Map, c(f = cbind, mget(ls(pattern = 'model_'))))
  
  export_df1 &lt;- Map(cbind, export_df, lst1)
  
  
  export_df1 &lt;- bind_rows(export_df1, .id = &quot;date&quot;)
  </code></pre>
  <p>I've deleted the other edits to save a lot more space.</p>
  <p>Edit2:</p>
  <pre><code>export_df1[[6]]
  [[1]]
        Jan      Feb      Mar      Apr      May      Jun      Jul      Aug      
  Sep      Oct      Nov      Dec
  2021 6.508872 4.639274 4.678671 4.626766 5.327353 6.890269 6.640483 
  6.164311 5.317675 6.152747 5.963053 5.243159
  2022 6.517052 4.645104 4.684550 4.632579 5.334046 6.898923 6.648823 
  6.172052 5.324353 6.160472 5.970540 5.249740
  
      export_df1[[6]]
  [[2]]
        Jan      Feb      Mar      Apr      May      Jun      Jul      Aug      
  Sep      Oct      Nov      Dec
  2021 6.508872 4.639274 4.678671 4.626766 5.327353 6.890269 6.640483 
  6.164311 5.317675 6.152747 5.963053 5.243159
  2022 6.517052 4.645104 4.684550 4.632579 5.334046 6.898923 6.648823 
  6.172052 5.324353 6.160472 5.970540 5.249740
  </code></pre>
  <p>Instead of having each row in my dataframe as a list of all of those values, I want the column in the form like below.</p>
  <pre><code>export_df1$actuals
    [1] 4.729682 4.573595 7.338069 4.742559 5.853501 3.167612 6.305137 
  5.879437 4.571004 5.367115 5.127305 4.552428 5.843000 4.060205 4.125869 
  4.190687
  [17] 4.894595 6.454123 4.686262 4.196005 7.289879 6.206924 6.191610 
  </code></pre>
  <p>6.100175 4.769656 4.829249 5.287280 4.425435 5.158180 4.402173 0.000000
  0.000000 etc</p>