is there a way how to custom format ridiculously large numbers (at least up to 10^100 in both ways) in google sheets:
thousands > K
millions > M
billions > B
trillions > T
etc...
negative quadrillions > Q
decillions > D
either via:
internal custom number formatting
formula (array formula ofc)
script similar to this one just extended to cover more ground
10000.1 10.0K
100 100.0
1000 1.0K
10000 10.0K
-100000 -100.0K
45646454 45.6M
5654894844216 5.7T
4655454544 4.7B
46546465455511 46.5T
-46546465455511 -46.5T
4654646545551184854556546454454400000000000000000000000000010000000 4.7U
-1000.9999 -1.0K
-100.8989 -100.9
-20.354 -20.4
1.03 1.0
22E+32 2.2D
internal custom number formatting solution:
sadly, the internal formatting in google sheets is by default able to work with only 3 types of numbers:
positive (1, 2, 5, 10, ...)
negative (-3, -9, -7, ...)
zero (0)
this can be tweaked to show custom formatting like thousands K, millions M and regular small numbers:
[>999999]0.0,,"M";[>999]0.0,"K";0
or only thousands K, millions M, billions B
[<999950]0.0,"K";[<999950000]0.0,,"M";0.0,,,"B"
or only negative thousands K, negative millions M, negative billions B
[>-999950]0.0,"K";[>-999950000]0.0,,"M";0.0,,,"B"
or only millions M, billions B, trillions T:
[<999950000]0.0,,"M";[<999950000000]0.0,,,"B";0.0,,,,"T"
or only numbers from negative million M to positive million M:
[>=999950]0.0,,"M";[<=-999950]0.0,,"M";0.0,"K"
but you always got only 3 slots you can use, meaning that you can't have trillions as the 4th type/slot. fyi, the 4th slot exists, but it's reserved for text. to learn more about internal formatting in google sheets see:
https://developers.google.com/sheets/api/guides/formats#meta_instructions
https://www.benlcollins.com/spreadsheets/google-sheets-custom-number-format/
formula (array formula) solution:
the formula approach is more versatile... first, you will need to decide on the system/standard you want to use (American, European, Greek, International, Unofficial, etc...):
en.wikipedia.org/wiki/Names_of_large_numbers
en.wikipedia.org/wiki/Metric_prefix
simple.wikipedia.org/wiki/Names_for_large_numbers
home.kpn.nl/vanadovv/BignumbyN
after that try:
=INDEX(REGEXREPLACE(IFNA(TEXT(A:A/10^(VLOOKUP(LEN(TEXT(INT(ABS(A:A)), "0"))-1,
SEQUENCE(35, 1,, 3), 1, 1)), "#.0")&VLOOKUP(ABS(A:A)*1, {{10^SEQUENCE(34, 1, 3, 3)},
{"K "; "M "; "B "; "T "; "Qa "; "Qi "; "Sx "; "Sp "; "O "; "N "; "D "; "Ud ";
"Dd "; "Td "; "Qad"; "Qid"; "Sxd"; "Spd"; "Od "; "Nd "; "V "; "Uv "; "Dv "; "Tv ";
"Qav"; "Qiv"; "Sxv"; "Spv"; "Ov "; "Nv "; "Tr "; "Ut "; "Dt "; "Tt "}}, 2, 1),
IF(ISBLANK(A:A),, TEXT(A:A, "0.0 "))), "^0\.0 $", "0 "))
works with positive numbers
works with negative numbers
works with zero
works with decimal numbers
works with numeric values
works with plain text numbers
works with scientific notations
works with blank cells
works up to googol 10^104 in both ways
extra points if you are interested in how it works...
let's start with virtual array {{},{}}. SEQUENCE(34, 1, 3, 3) will give us 34 numbers in 1 column starting from number 3 with the step of 3 numbers:
these will be used as exponents while rising 10 on the power ^
so our virtual array will be:
next, we insert it as the 2nd argument of VLOOKUP where we check ABS absolute values (converting negative values into positive) of A column multiplied by *1 just in case values of A column are not numeric. via VLOOKUP we return the second 2 column and as the 4th argument, we use approximate mode 1
numbers from -999 to 999 will intentionally error out at this point so we could later use IFNA to "fix" our errors with IF(A:A=IF(,,),, TEXT(A:A, "#.0 ")) translated as: if range A:A is truly empty ISBLANK output nothing, else format A column with provided pattern #.0 eg. if cell A5 = empty, the output will be blank cell... if -999 < A5=50 < 999 the output will be 50.0
and the last part:
TEXT(A:A/10^(VLOOKUP(LEN(TEXT(INT(ABS(A:A)), "0"))-1,
SEQUENCE(35, 1,, 3), 1, 1)), "#.0")
ABS(A:A) to convert negative numbers into positive. INT to remove decimal numbers if any. TEXT(, "0") to convert scientific notations 3E+8 into regular numbers 300000000. LEN to count digits. -1 to correct for base10 notation. VLOOKUP above-constructed number in SEQUENCE of 35 numbers in 1 column, this time starting from number 0 ,, with the step of 3 numbers. return via VLOOKUP the first 1 column (eg. the sequence) in approximate mode 1 of vlookup. insert this number as exponent when rising the 10 on power ^. and take values in A column and divide it by the above-constructed number 10 raised on the power ^ of a specific exponent. and lastly, format it with TEXT as #.0
to convert ugly 0.0 into beautiful 0 we just use REGEXREPLACE. and INDEX is used instead of the longer ARRAYFORMULA.
sidenote: to remove trailing spaces (which are there to add nice alignment lol) either remove them from the formula or use TRIM right after INDEX.
script solution:
gratitude to #TheMaster for covering this
here is a mod of it:
/**
* formats various numbers according to the provided short format
* #customfunction
* #param {A1:C100} range a 2D array
* #param {[X1:Y10]} database [optional] a real/virtual 2D array
* where the odd column holds exponent of base 10
* and the even column contains format suffixes
* #param {[5]} value [optional] fix suffix to fixed length
* by padding spaces (only if the second parameter exists)
*/
// examples:
// =CSF(A1:A)
// =CSF(2:2; X5:Y10)
// =CSF(A1:3; G10:J30)
// =CSF(C:C; X:Y; 2) to use custom alignment
// =CSF(C:C; X:Y; 0) to remove alignment
// =INDEX(TRIM(CSF(A:A))) to remove alignment
// =CSF(B10:D30; {3\ "K"; 4\ "TK"}) for non-english sheets
// =CSF(E5, {2, "deci"; 3, "kilo"}) for english sheets
// =INDEX(IF(ISERR(A:A*1); A:A; CSF(A:A))) to return non-numbers
// =INDEX(IF((ISERR(A:A*1))+(ISBLANK(A:A)), A:A, CSF(A:A*1))) enforce mode
function CSF(
range,
database = [
[3, 'K' ], //Thousand
[6, 'M' ], //Million
[9, 'B' ], //Billion
[12, 'T' ], //Trillion
[15, 'Qa' ], //Quadrillion
[18, 'Qi' ], //Quintillion
[21, 'Sx' ], //Sextillion
[24, 'Sp' ], //Septillion
[27, 'O' ], //Octillion
[30, 'N' ], //Nonillion
[33, 'D' ], //Decillion
[36, 'Ud' ], //Undecillion
[39, 'Dd' ], //Duodecillion
[42, 'Td' ], //Tredecillion
[45, 'Qad'], //Quattuordecillion
[48, 'Qid'], //Quindecillion
[51, 'Sxd'], //Sexdecillion
[54, 'Spd'], //Septendecillion
[57, 'Od' ], //Octodecillion
[60, 'Nd' ], //Novemdecillion
[63, 'V' ], //Vigintillion
[66, 'Uv' ], //Unvigintillion
[69, 'Dv' ], //Duovigintillion
[72, 'Tv' ], //Trevigintillion
[75, 'Qav'], //Quattuorvigintillion
[78, 'Qiv'], //Quinvigintillion
[81, 'Sxv'], //Sexvigintillion
[84, 'Spv'], //Septenvigintillion
[87, 'Ov' ], //Octovigintillion
[90, 'Nv' ], //Novemvigintillion
[93, 'Tr' ], //Trigintillion
[96, 'Ut' ], //Untrigintillion
[99, 'Dt' ], //Duotrigintillion
[100, 'G' ], //Googol
[102, 'Tt' ], //Tretrigintillion or One Hundred Googol
],
value = 3
) {
if (
database[database.length - 1] &&
database[database.length - 1][0] !== 0
) {
database = database.reverse();
database.push([0, '']);
}
const addSuffix = num => {
const pad3 = (str = '') => str.padEnd(value, ' ');
const decim = 1 // round to decimal places
const separ = 0 // separate number and suffix
const anum = Math.abs(num);
if (num === 0)
return '0' + ' ' + ' '.repeat(separ) + ' '.repeat(decim) + pad3();
if (anum > 0 && anum < 1)
return String(num.toFixed(decim)) + ' '.repeat(separ) + pad3();
for (const [exp, suffix] of database) {
if (anum >= Math.pow(10, exp))
return `${(num / Math.pow(10, exp)).toFixed(decim)
}${' '.repeat(separ) + pad3(suffix)}`;
}
};
return customFunctionRecurse_(
range, CSF, addSuffix, database, value, true
);
}
function customFunctionRecurse_(
array, mainFunc, subFunc, ...extraArgToMainFunc
) {
if (Array.isArray(array))
return array.map(e => mainFunc(e, ...extraArgToMainFunc));
else return subFunc(array);
}
sidenote 1: this script does not need to be authorized priorly to usage
sidenote 2: cell formatting needs to be set to Automatic or Number otherwise use enforce mode
extra:
convert numbers into plain text strings/words
convert array of numbers into plain text strings/words
convert custom formatted numbers into numeric numbers/values
convert text string datetime into duration value
convert text string formatted numbers into duration
convert your age into years-months-days
For almost all practical purposes we can use Intl compact format to achieve this functionality.
/**
* Utility function needed to recurse 2D arrays
*/
function customFunctionRecurse_(
array,
mainFunc,
subFunc,
...extraArgToMainFunc
) {
if (Array.isArray(array))
return array.map(e => mainFunc(e, ...extraArgToMainFunc));
else return subFunc(array);
}
/**
* Simple custom formating function using Intl
* #see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat
* #customfunction
* #author TheMaster https://stackoverflow.com/users/8404453
* #param {A1:D2} numArr A 2D array
* #returns {String[][]}Compact Intl formatted 2D array
*/
function format(numArr) {
const cIntl = new Intl.NumberFormat('en-GB', {
notation: 'compact',
compactDisplay: 'short',
});
return customFunctionRecurse_(numArr, format, (num) => cIntl.format(num));
}
But for extreme ends or custom formatting, We need to use a custom script:
/**
* Formats various numbers according to the provided format
* #customfunction
* #author TheMaster https://stackoverflow.com/users/8404453
* #param {A1:D2} numArr A 2D array
* #param {X1:Y2} formatArr [optional] A format 2D real/virtual array
* with base 10 power -> suffix mapping
* eg: X1:3 Y1:K represents numbers > 10^3 should have a K suffix
* #param {3} suffixPadLength [optional] Fix suffix to fixed length by padding spaces
* #returns {String[][]} Formatted 2D array
*/
function customFormat(
numArr,
formatArr = [
/**This formatArr array is provided by
* by player0 https://stackoverflow.com/users/5632629/
* #see https://stackoverflow.com/questions/69773823#comment123503634_69809210
*/
[3, 'K'], //Thousand
[6, 'M'], //Million
[9, 'B'], //Billion
[12, 'T'], //Trillion
[15, 'Qa'], //Quadrillion
[18, 'Qi'], //Quintillion
[21, 'Sx'], //Sextillion
[24, 'Sp'], //Septillion
[27, 'O'], //Octillion
[30, 'N'], //Nonillion
[33, 'D'], //Decillion
[36, 'Ud'], //Undecillion
[39, 'Dd'], //Duodecillion
[42, 'Td'], //Tredecillion
[45, 'Qad'], //Quattuordecillion
[48, 'Qid'], //Quindecillion
[51, 'Sxd'], //Sexdecillion
[54, 'Spd'], //Septendecillion
[57, 'Od'], //Octodecillion
[60, 'Nd'], //Novemdecillion
[63, 'V'], //Vigintillion
[66, 'Uv'], //Unvigintillion
[69, 'Dv'], //Duovigintillion
[72, 'Tv'], //Trevigintillion
[75, 'Qav'], //Quattuorvigintillion
[78, 'Qiv'], //Quinvigintillion
[81, 'Sxv'], //Sexvigintillion
[84, 'Spv'], //Septenvigintillion
[87, 'Ov'], //Octovigintillion
[90, 'Nv'], //Novemvigintillion
[93, 'Tr'], //Trigintillion
[96, 'Ut'], //Untrigintillion
[99, 'Dt'], //Duotrigintillion
[102, 'G'], //Googol
],
suffixPadLength = 3,
inRecursion = false
) {
if (!inRecursion) {
formatArr = formatArr.reverse();
formatArr.push([0, '']);
}
const addSuffix = num => {
const pad3 = (str = '') => str.padEnd(suffixPadLength, ' '); //pad 3 spaces if necessary
const anum = Math.abs(num);
if (num === 0) return '0' + pad3();
if (anum > 0 && anum < 1) return String(num.toFixed(2)) + pad3();
for (const [exp, suffix] of formatArr) {
if (anum >= Math.pow(10, exp))
return `${(num / Math.pow(10, exp)).toFixed(2)}${pad3(suffix)}`;
}
};
return customFunctionRecurse_(
numArr,
customFormat,
addSuffix,
formatArr,
suffixPadLength,
true
);
}
Usage:
=CUSTOMFORMAT(A1:A5,{{3,"k"};{10,"G"}})
Tells custom function to use k for numbers>10^3 and G for 10^10
Illustration:
/*<ignore>*/console.config({maximize:true,timeStamps:false,autoScroll:false});/*</ignore>*/
/**
* Utility function needed to map 2D arrays
*/
function customFunctionRecurse_(array, mainFunc, subFunc, extraArgToMainFunc) {
if (Array.isArray(array))
return array.map((e) => mainFunc(e, extraArgToMainFunc));
else return subFunc(array);
}
/**
* Simple custom formating function using Intl
* #see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat
* #customfunction
* #param {A1:D2} A 2D array
* #returns {String[][]}Compact Intl formatted 2D array
*/
function format(numArr) {
const cIntl = new Intl.NumberFormat('en-GB', {
notation: 'compact',
compactDisplay: 'short',
});
return customFunctionRecurse_(numArr, format, (num) => cIntl.format(num));
}
/**
* Formats various numbers according to the provided format
* #customfunction
* #param {A1:D2} A 2D array
* #param {X1:Y2=} [optional] A format 2D real/virtual array
* with base 10 power -> suffix mapping
* eg: X1:3 Y1:K represents numbers > 10^3 should have a K suffix
* #returns {String[][]} Formatted 2D array
*/
function customFormat(
numArr,
formatArr = [
//sample byte => kb formatting
[3, 'kb'],
[6, 'mb'],
[9, 'gb'],
[12, 'tb'],
]
) {
//console.log({ numArr, formatArr });
if (
formatArr[formatArr.length - 1] &&
formatArr[formatArr.length - 1][0] !== 0
) {
formatArr = formatArr.reverse();
formatArr.push([0, '']);
}
const addSuffix = (num) => {
const anum = Math.abs(num);
if (num === 0) return '0.00';
if (anum > 0 && anum < 1) return String(num.toFixed(2));
for (const [exp, suffix] of formatArr) {
if (anum >= Math.pow(10, exp))
return `${(num / Math.pow(10, exp)).toFixed(2)}${suffix}`;
}
};
return customFunctionRecurse_(numArr, customFormat, addSuffix, formatArr);
}
console.log(
customFormat([
[
0,
1000,
153,
12883255,
235688235123,
88555552233355888,
-86555,
0.8523588055,
Math.pow(10, 15),
],
])
);
<!-- https://meta.stackoverflow.com/a/375985/ --> <script src="https://gh-canon.github.io/stack-snippet-console/console.min.js"></script>
sometimes when we deal with nuclear physics we need to shorten the time so this is how:
=INDEX(IF(ISBLANK(A2:A),,TEXT(TRUNC(TEXT(IF(A2:A*1<1,
TEXT(A2:A*1, "0."&REPT(0, 30))*VLOOKUP(A2:A*1, {SORT({0; 1/10^SEQUENCE(9, 1, 3, 3)}),
{0; 10^SORT(SEQUENCE(9, 1, 3, 3), 1,)}}, 2, 1), TEXT(A2:A*1, REPT(0, 30))/
VLOOKUP(A2:A*1, TEXT({1; 60; 3600; 86400; 31536000; 31536000*10^SEQUENCE(8, 1, 3, 3)},
{"#", "#"})*1, 2, 1)), "0."&REPT("0", 30)), 3), "0.000")&" "&
VLOOKUP(A2:A*1, {SORT({0; 1/10^SEQUENCE(9, 1, 3, 3);
{1; 60; 3600; 86400; 31536000}; 31536000*10^SEQUENCE(8, 1, 3, 3)}), FLATTEN(SPLIT(
"s ys zs as fs ps ns μs ms s m h d y ky My Gy Ty Py Ey Zy Yy", " ",,))}, 2, 1)))
it's a simple conversion from seconds into abbreviation utilizing the International System of Units where:
in seconds
____________________________________
ys = yoctosecond = 0.000000000000000000000001
zs = zeptosecond = 0.000000000000000000001
as = attosecond = 0.000000000000000001
fs = femtosecond = 0.000000000000001
ps = pikosecond = 0.000000000001
ns = nanosecond = 0.000000001
μs = microsecond = 0.000001
ms = millisecond = 0.001
s = second = 1
m = minute = 60
h = hour = 3600
d = day = 86400
y = year = 31536000
ky = kiloyear = 31536000000
My = megayear = 31536000000000
Gy = gigayear = 31536000000000000
Ty = terayear = 31536000000000000000
Py = petayear = 31536000000000000000000
Ey = exayear = 31536000000000000000000000
Zy = zettayear = 31536000000000000000000000000
Yy = yottayear = 31536000000000000000000000000000
I am working on a simple 2D peg-in-hole example in which I want to directly control the peg (which is just a box) in 2D. In order to do that, I add a planar joint to attach the peg to the world as in the Manipulation course's example.
planar_joint_frame = plant.AddFrame(FixedOffsetFrame("planar_joint_frame", plant.world_frame(),RigidTransform(RotationMatrix.MakeXRotation(np.pi/2))))
peg_planar_joint_frame = plant.AddFrame(FixedOffsetFrame("peg_planar_joint_frame", peg.body_frame(),RigidTransform(RotationMatrix.MakeXRotation(np.pi/2), [0, 0, 0.025])))
peg_planar_joint = plant.AddJoint(PlanarJoint("peg_planar_joint", planar_joint_frame,peg_planar_joint_frame, damping=[0, 0, 0]))
Then I want to add actuator to the peg_planar_joint so that I can directly apply planar force fx, fy and torque tau_z.
plant.AddJointActuator("peg_planar_joint", peg_planar_joint)
However, I encounter the following error
SystemExit: Failure at bazel-out/k8-opt/bin/multibody/plant/_virtual_includes/multibody_plant_core/drake/multibody/plant/multibody_plant.h:1131 in AddJointActuator(): condition 'joint.num_velocities() == 1' failed.
It seems that JointActuator can only be added to PrismaticJoint or RevoluteJoint that is with 1 degree of freedom. So my question is can I add JointActuator seperately to the x, y and RotZ of PlanarJoint? If not, what is the workaround method? Thank you!
Right now you have to add the prismatic and revolute joint's yourself to add actuators. Here is a snippet from https://github.com/RussTedrake/manipulation/blob/f868cd684a35ada15d6063c70daf1c9d61000941/force.ipynb
# Add a planar joint the old fashioned way (so that I can have three actuators):
gripper_false_body1 = plant.AddRigidBody(
"false_body1", gripper,
SpatialInertia(0, [0, 0, 0], UnitInertia(0, 0, 0)))
gripper_false_body2 = plant.AddRigidBody(
"false_body2", gripper,
SpatialInertia(0, [0, 0, 0], UnitInertia(0, 0, 0)))
gripper_x = plant.AddJoint(
PrismaticJoint("gripper_x", plant.world_frame(),
plant.GetFrameByName("false_body1"), [1, 0, 0], -.3, .3))
plant.AddJointActuator("gripper_x", gripper_x)
gripper_z = plant.AddJoint(
PrismaticJoint("gripper_z", plant.GetFrameByName("false_body1"),
plant.GetFrameByName("false_body2"), [0, 0, 1], 0.0,
0.5))
gripper_z.set_default_translation(0.3)
plant.AddJointActuator("gripper_z", gripper_z)
gripper_frame = plant.AddFrame(
FixedOffsetFrame(
"gripper_frame", plant.GetFrameByName("body", gripper),
RigidTransform(RotationMatrix.MakeXRotation(np.pi / 2))))
gripper_theta = plant.AddJoint(
RevoluteJoint("gripper_theta", plant.GetFrameByName("false_body2"),
gripper_frame, [0, -1, 0], -np.pi, np.pi))
plant.AddJointActuator("gripper_theta", gripper_theta)
It depends on what you want to accomplish really. If all you need is the sim of a 2D system with externally applied forces, then I'd say you look into MultibodyPlant::get_applied_generalized_force_input_port().
I have a piece of code which I use to visualize a graph:
if (visualize == True):
# Black removed and is used for noise instead.
unique_labels = set(db.labels_)
colors = [plt.cm.Spectral(each)
for each in np.linspace(0, 1, len(unique_labels))]
for k, col in zip(unique_labels, colors):
if k == -1:
# Black used for noise.
col = [0, 0, 0, 1]
class_member_mask = (db.labels_ == k)
xy = scaled_points[class_member_mask & core_samples_mask]
plt.plot(xy[:, 0], xy[:, 1], 'o', markerfacecolor=tuple(col),
markeredgecolor='k', markersize=14)
xy = scaled_points[class_member_mask & ~core_samples_mask]
plt.plot(xy[:, 0], xy[:, 1], 'o', markerfacecolor=tuple(col),
markeredgecolor='k', markersize=6)
# display the graph
plt.title('Estimated number of clusters: %d' % n_clusters_)
plt.show()
# get the image into a variable that OpenCV likes
# uh, how?
while this works, I want to have the end result (whatever is being shown) as an OpenCV image.
Since I don't even have the variable -image-, I have no idea how to achieve this.
Did anyone do something similar?
EDIT: I am actually getting close. Now I can create an OpenCV image out of a fig, but the contents are not right. The fig is empty. I wonder where I go wrong? Why doesn't it get the plt object from above and draw the actual content?
fig = plt.figure()
canvas = FigureCanvas(fig)
canvas.draw()
# convert canvas to image
graph_image = np.fromstring(fig.canvas.tostring_rgb(), dtype=np.uint8, sep='')
graph_image = graph_image.reshape(fig.canvas.get_width_height()[::-1] + (3,))
# it still is rgb, convert to opencv's default bgr
graph_image = cv2.cvtColor(graph_image,cv2.COLOR_RGB2BGR)
Okay, I finally got it! One has to create the fig object at the very beginning, then use the necessary plotting functions, then convert to canvas and then to OpenCV image.
EDIT: Thanks to the suggestion of #ImportanceOfBeingErnest, now the code is even more straightforward!
Here is the full code:
if (visualize == True):
# create a figure
fig = plt.figure()
# Black removed and is used for noise instead.
unique_labels = set(db.labels_)
colors = [plt.cm.Spectral(each)
for each in np.linspace(0, 1, len(unique_labels))]
for k, col in zip(unique_labels, colors):
if k == -1:
# Black used for noise.
col = [0, 0, 0, 1]
class_member_mask = (db.labels_ == k)
xy = scaled_points[class_member_mask & core_samples_mask]
plt.plot(xy[:, 0], xy[:, 1], 'o', markerfacecolor=tuple(col),
markeredgecolor='k', markersize=14)
xy = scaled_points[class_member_mask & ~core_samples_mask]
plt.plot(xy[:, 0], xy[:, 1], 'o', markerfacecolor=tuple(col),
markeredgecolor='k', markersize=6)
# convert it to an OpenCV image/numpy array
canvas = FigureCanvas(fig)
canvas.draw()
# convert canvas to image
graph_image = np.array(fig.canvas.get_renderer()._renderer)
# it still is rgb, convert to opencv's default bgr
graph_image = cv2.cvtColor(graph_image,cv2.COLOR_RGB2BGR)