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Formatting a table to look nice

How to create a Table of Tables with indexed variablesTable shifting bug?Transposing a tableHow to create a table of tables with different table lengths?Looping with “Table” over two variablesHow to plot data from table dynamically, without knowing how many columns are there?Creating iterations on a circle using the Table functionProgress bar / counter for multi-row table with 2 variablesPlotting in Table[…]Attempting to fill a table with the number of elements in each bin and make a table with the elements in the bins?

7

1

$begingroup$

my current code is:

binsize = 21;

data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 

   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 

   477, 518, 538, 497, 565};

firstbin = 472;

a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];

b = Range[firstbin, Max[data] + binsize, binsize];

Transpose[{Take[b, Length[a]], a}] // TableForm

Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"

any advice would be great, thank you.

table formatting

share|improve this question

edited 22 hours ago

Wombles

asked 22 hours ago

WomblesWombles

534

$endgroup$

add a comment | 

7

1

$begingroup$

my current code is:

binsize = 21;

data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 

   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 

   477, 518, 538, 497, 565};

firstbin = 472;

a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];

b = Range[firstbin, Max[data] + binsize, binsize];

Transpose[{Take[b, Length[a]], a}] // TableForm

Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"

any advice would be great, thank you.

table formatting

share|improve this question

edited 22 hours ago

Wombles

asked 22 hours ago

WomblesWombles

534

$endgroup$

add a comment | 

$begingroup$

my current code is:

binsize = 21;

data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 

   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 

   477, 518, 538, 497, 565};

firstbin = 472;

a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];

b = Range[firstbin, Max[data] + binsize, binsize];

Transpose[{Take[b, Length[a]], a}] // TableForm

Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"

any advice would be great, thank you.

table formatting

share|improve this question

edited 22 hours ago

Wombles

asked 22 hours ago

WomblesWombles

534

$endgroup$

binsize = 21;

data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 

   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 

   477, 518, 538, 497, 565};

firstbin = 472;

a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];

b = Range[firstbin, Max[data] + binsize, binsize];

Transpose[{Take[b, Length[a]], a}] // TableForm

share|improve this question

edited 22 hours ago

Wombles

asked 22 hours ago

WomblesWombles

534

share|improve this question

edited 22 hours ago

Wombles

asked 22 hours ago

WomblesWombles

534

asked 22 hours ago

WomblesWombles

534

asked 22 hours ago

WomblesWombles

534

2 Answers
2

active

oldest

votes

8

$begingroup$

You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:

{binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];

bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];

TableForm[Transpose[{bins, bincounts}]]

Alternatively, you can use MovingMap, Developer`PartitionMap or
the (undocumented) 6-argument form of Partition to get the first column:

bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]

bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];

bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];

bins == bins2 == bins3 == bins4

True

Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:

bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]

TableForm[Transpose[{bins5, bincounts}]]

share|improve this answer

edited 16 hours ago

answered 22 hours ago

kglrkglr

187k10203421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – Wombles
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wombles, you are welcome.
  $endgroup$
  – kglr
  21 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Here is one way:

bb = b - 1;

c = Complement[bb, {Min[bb]}];

Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm

Here is the output:

With MarcoB's hint, and some experimenting:

c = Complement[b - 1, {Min[b - 1]}];

y = Map[ToString, Take[b, Length[a]]];

z = Map[ToString, c]; 

Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm

share|improve this answer

edited 16 hours ago

answered 22 hours ago

mjwmjw

4828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
  $endgroup$
  – mjw
  22 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You might want ToString and StringJoin
  $endgroup$
  – MarcoB
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
  $endgroup$
  – mjw
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
  $endgroup$
  – mjw
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

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8

$begingroup$

You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:

{binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];

bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];

TableForm[Transpose[{bins, bincounts}]]

Alternatively, you can use MovingMap, Developer`PartitionMap or
the (undocumented) 6-argument form of Partition to get the first column:

bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]

bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];

bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];

bins == bins2 == bins3 == bins4

True

Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:

bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]

TableForm[Transpose[{bins5, bincounts}]]

share|improve this answer

edited 16 hours ago

answered 22 hours ago

kglrkglr

187k10203421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – Wombles
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wombles, you are welcome.
  $endgroup$
  – kglr
  21 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:

{binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];

bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];

TableForm[Transpose[{bins, bincounts}]]

Alternatively, you can use MovingMap, Developer`PartitionMap or
the (undocumented) 6-argument form of Partition to get the first column:

bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]

bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];

bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];

bins == bins2 == bins3 == bins4

True

Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:

bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]

TableForm[Transpose[{bins5, bincounts}]]

share|improve this answer

edited 16 hours ago

answered 22 hours ago

kglrkglr

187k10203421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – Wombles
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wombles, you are welcome.
  $endgroup$
  – kglr
  21 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:

{binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];

bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];

TableForm[Transpose[{bins, bincounts}]]

Alternatively, you can use MovingMap, Developer`PartitionMap or
the (undocumented) 6-argument form of Partition to get the first column:

bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]

bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];

bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];

bins == bins2 == bins3 == bins4

True

Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:

bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]

TableForm[Transpose[{bins5, bincounts}]]

share|improve this answer

edited 16 hours ago

answered 22 hours ago

kglrkglr

187k10203421

$endgroup$

{binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];

bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];

TableForm[Transpose[{bins, bincounts}]]

bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]

bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];

bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];

bins == bins2 == bins3 == bins4

bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]

TableForm[Transpose[{bins5, bincounts}]]

share|improve this answer

edited 16 hours ago

answered 22 hours ago

kglrkglr

187k10203421

answered 22 hours ago

kglrkglr

187k10203421

answered 22 hours ago

kglrkglr

187k10203421

4

$begingroup$

Here is one way:

bb = b - 1;

c = Complement[bb, {Min[bb]}];

Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm

Here is the output:

With MarcoB's hint, and some experimenting:

c = Complement[b - 1, {Min[b - 1]}];

y = Map[ToString, Take[b, Length[a]]];

z = Map[ToString, c]; 

Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm

share|improve this answer

edited 16 hours ago

answered 22 hours ago

mjwmjw

4828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
  $endgroup$
  – mjw
  22 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You might want ToString and StringJoin
  $endgroup$
  – MarcoB
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
  $endgroup$
  – mjw
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
  $endgroup$
  – mjw
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Here is one way:

bb = b - 1;

c = Complement[bb, {Min[bb]}];

Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm

Here is the output:

With MarcoB's hint, and some experimenting:

c = Complement[b - 1, {Min[b - 1]}];

y = Map[ToString, Take[b, Length[a]]];

z = Map[ToString, c]; 

Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm

share|improve this answer

edited 16 hours ago

answered 22 hours ago

mjwmjw

4828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
  $endgroup$
  – mjw
  22 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You might want ToString and StringJoin
  $endgroup$
  – MarcoB
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
  $endgroup$
  – mjw
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
  $endgroup$
  – mjw
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Here is one way:

bb = b - 1;

c = Complement[bb, {Min[bb]}];

Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm

Here is the output:

With MarcoB's hint, and some experimenting:

c = Complement[b - 1, {Min[b - 1]}];

y = Map[ToString, Take[b, Length[a]]];

z = Map[ToString, c]; 

Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm

share|improve this answer

edited 16 hours ago

answered 22 hours ago

mjwmjw

4828

$endgroup$

bb = b - 1;

c = Complement[bb, {Min[bb]}];

Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm

c = Complement[b - 1, {Min[b - 1]}];

y = Map[ToString, Take[b, Length[a]]];

z = Map[ToString, c]; 

Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm

share|improve this answer

edited 16 hours ago

answered 22 hours ago

mjwmjw

4828

answered 22 hours ago

mjwmjw

4828

answered 22 hours ago

mjwmjw

4828