# How can I create a row of numbers form 1 to 10 ?. like this [1.0000, 2.0000, 3.0000,...]?

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Chinonso Collins Unaichi
on 3 May 2016

Commented: Chinonso Collins Unaichi
on 3 May 2016

##### 0 Comments

### Accepted Answer

John BG
on 3 May 2016

with

format long;x=1:10

=

1 2 3 4 5 6 7 8 9 10

vpa only adds one zero:

y=vpa(x,5)

=

[ 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0]

y=vpa(x,10)

=

[ 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0]

floating point boundaries (without vpa):

flintmax

=

9.007199254740992e+15

intmax

=

2147483647

intmin

=

-2147483648

with fixed point precision objects, details here: http://uk.mathworks.com/help/fixedpoint/ref/fi.html?searchHighlight=fi

you get more than 4 zeros:

a = fi(pi, 1, 8, 3)

a =

3.125000000000000

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 8

FractionLength: 3

but only if it is rounding non-zero decimals:

a = fi(1, 1, 8, 3)

=

1

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 8

FractionLength: 3

a = fi(1.0000, 1, 8, 3)

=

1

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 8

FractionLength: 3

tried 1.0001 and then -0001 but did not work:

b = fi(1.0001, 1) b = 1.000122070312500

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 16

FractionLength: 14

b-.0001

Warning: The behavior of adding or subtracting a fi and a non-fi operand has

changed. The non-fi operand is now cast to the data type of the fi operand.

See the release notes for information on retaining the old behavior of best

precision scaling cast. In - (line 37)

=

1

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 17

FractionLength: 14

c=b-.0001 = 1

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 17

FractionLength: 14

b2=fi(0.0001,1)

=

1.000016927719116e-04

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 16

FractionLength: 28

b-b2 = 1.000022068619728

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 31

FractionLength: 28

d=b-b2 = 1.000022068619728

DataTypeMode: Fixed-point: binary point scaling

Signedness: Signed

WordLength: 31

FractionLength: 28

d2=vpa(d,4)

=

1.0

If you really want to present the vector exactly with 4 zero decimals, you can translate each figure to char,

x=[1:1:4]

x_char=char(x+48)'

=

1

2

3

4

and then add '.000'

x2=repmat(['.0000'],4,1)

=

.0000

.0000

.0000

.0000

x_char=[x_char x2]

=

1.0000

2.0000

3.0000

4.0000

If you find this answer of any help solving your question,

please click on the thumbs-up vote link,

thanks in advance

John

### More Answers (1)

Alessandro Masullo
on 3 May 2016

Matlab deals with double variables by default. If you write 1, it is actually 1.00000000

If you just want to display it, you can use sprintf('%.5f',1).

To check your variable type, you can use whos

##### 2 Comments

Guillaume
on 3 May 2016

Edited: Guillaume
on 3 May 2016

To add to Alessandro's answer. You're confusing two different things, the number that is stored in the matrix and the way it is displayed.

In matlab and in math, there is no difference whatsoever between 1 and 1.0000. You can change the way matlab displays number in the command window with format. Most of the formats display integer without any decimal, only the engineering format always use decimal:

>>format shorteng

>>1

ans =

1.0000e+000

Or as Alessandro said, you can use the format string of sprintf to output your number to the command line.

But as said, it is just cosmetics. The numbers are the same regardless of how you display them.

Alessandro Masullo
on 3 May 2016

Yes, you're right. The only real difference comes when you specify the variable type:

K>> format shorteng

K>> double(1:5)

ans =

1.0000e+000 2.0000e+000 3.0000e+000 4.0000e+000 5.0000e+000

K>> uint8(1:5)

ans =

1 2 3 4 5

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