add another way to extract row as a vector

Author: bkamins

06_rows.ipynb

@@ -5,7 +5,7 @@
    "metadata": {},
    "source": [
     "# Introduction to DataFrames\n",
-    "**[Bogumił Kamiński](http://bogumilkaminski.pl/about/), Apr 21, 2018**"
+    "**[Bogumił Kamiński](http://bogumilkaminski.pl/about/), Jun 2, 2018**"
    ]
   },
   {
@@ -1195,6 +1195,28 @@
    "source": [
     "Tuple(x[1, col] for col in cols) # similar construct for Tuples, when ported to Julia 0.7 NamedTuples will be added"
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "2-element Array{Any,1}:\n",
+       " 1   \n",
+       "  \"a\""
+      ]
+     },
+     "execution_count": 40,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "vec(Matrix(x[1, cols])) # also you can use a conversion to a Matrix"
+   ]
   }
  ],
  "metadata": {
@@ -1207,7 +1229,7 @@
    "file_extension": ".jl",
    "mimetype": "application/julia",
    "name": "julia",
-   "version": "0.6.2"
+   "version": "0.6.3"
   }
  },
  "nbformat": 4,

README.md

@@ -77,3 +77,5 @@ Changelog:
 
 1. Improved rendering of `#undef` in HTML/LaTeX.
 2. Added `permutecols!` function.
+3. `describe` returns a `DataFrame`
+4. On Julia 0.7 you can access columns of `DataFrame` using `.` notation