Formatting a table to look nice
5
$begingroup$
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
asked 9 hours ago
WomblesWombles
434
$endgroup$
add a comment |
5
$begingroup$
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
asked 9 hours ago
WomblesWombles
434
$endgroup$
add a comment |
5
5
5
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
asked 9 hours ago
WomblesWombles
434
$endgroup$
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
table formatting
asked 9 hours ago
WomblesWombles
434
asked 9 hours ago
WomblesWombles
434
edited 9 hours ago
Wombles
asked 9 hours ago
WomblesWombles
434
asked 9 hours ago
WomblesWombles
434
asked 9 hours ago
WomblesWombles
434
434
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
7
$begingroup$
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}]]
answered 8 hours ago
kglrkglr
187k10203421
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
add a comment |
3
$begingroup$
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
answered 8 hours ago
mjwmjw
4428
$endgroup$
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
1
$begingroup$
You might wantToStringandStringJoin
$endgroup$
– MarcoB
8 hours ago
$begingroup$
@MarcoB, Yes, Thank you! That's what I was looking for! I actually foundToStringand 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
4 hours ago
$begingroup$
Rowas suggested in kglr's answer, can also work as a substitute forStringJoin.
$endgroup$
– mjw
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
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}]]
answered 8 hours ago
kglrkglr
187k10203421
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
add a comment |
7
$begingroup$
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}]]
answered 8 hours ago
kglrkglr
187k10203421
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
add a comment |
7
7
7
$begingroup$
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}]]
answered 8 hours ago
kglrkglr
187k10203421
$endgroup$
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}]]
answered 8 hours ago
kglrkglr
187k10203421
edited 3 hours ago
answered 8 hours ago
kglrkglr
187k10203421
answered 8 hours ago
kglrkglr
187k10203421
answered 8 hours ago
kglrkglr
187k10203421
187k10203421
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
Thank you very much!
$endgroup$
– Wombles
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
$begingroup$
@Wombles, you are welcome.
$endgroup$
– kglr
8 hours ago
add a comment |
3
$begingroup$
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
answered 8 hours ago
mjwmjw
4428
$endgroup$
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
1
$begingroup$
You might wantToStringandStringJoin
$endgroup$
– MarcoB
8 hours ago
$begingroup$
@MarcoB, Yes, Thank you! That's what I was looking for! I actually foundToStringand 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
4 hours ago
$begingroup$
Rowas suggested in kglr's answer, can also work as a substitute forStringJoin.
$endgroup$
– mjw
3 hours ago
add a comment |
3
$begingroup$
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
answered 8 hours ago
mjwmjw
4428
$endgroup$
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
1
$begingroup$
You might wantToStringandStringJoin
$endgroup$
– MarcoB
8 hours ago
$begingroup$
@MarcoB, Yes, Thank you! That's what I was looking for! I actually foundToStringand 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
4 hours ago
$begingroup$
Rowas suggested in kglr's answer, can also work as a substitute forStringJoin.
$endgroup$
– mjw
3 hours ago
add a comment |
3
3
3
$begingroup$
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
answered 8 hours ago
mjwmjw
4428
$endgroup$
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
answered 8 hours ago
mjwmjw
4428
edited 3 hours ago
answered 8 hours ago
mjwmjw
4428
answered 8 hours ago
mjwmjw
4428
answered 8 hours ago
mjwmjw
4428
4428
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
1
$begingroup$
You might wantToStringandStringJoin
$endgroup$
– MarcoB
8 hours ago
$begingroup$
@MarcoB, Yes, Thank you! That's what I was looking for! I actually foundToStringand 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
4 hours ago
$begingroup$
Rowas suggested in kglr's answer, can also work as a substitute forStringJoin.
$endgroup$
– mjw
3 hours ago
add a comment |
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
1
$begingroup$
You might wantToStringandStringJoin
$endgroup$
– MarcoB
8 hours ago
$begingroup$
@MarcoB, Yes, Thank you! That's what I was looking for! I actually foundToStringand 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
4 hours ago
$begingroup$
Rowas suggested in kglr's answer, can also work as a substitute forStringJoin.
$endgroup$
– mjw
3 hours ago
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
$begingroup$
I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
$endgroup$
– mjw
8 hours ago
1
1
$begingroup$
You might want
ToString and StringJoin$endgroup$
– MarcoB
8 hours ago
$begingroup$
You might want
ToString and StringJoin$endgroup$
– MarcoB
8 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
4 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
4 hours ago
$begingroup$
Row as suggested in kglr's answer, can also work as a substitute for StringJoin.$endgroup$
– mjw
3 hours ago
$begingroup$
Row as suggested in kglr's answer, can also work as a substitute for StringJoin.$endgroup$
– mjw
3 hours ago
add a comment |
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