CODEWITHSUNDEEP
r
m3k_1
m3k_1

Reputation: 417

cut() function puts all data in a single interval

I'm trying to cut a data to do a frequency distribution but after cut, all the data is assingned to one interval

points <- 224 * 0:5
cut_data <- cut(rs$amount, points ,dig.lab = 10)

My rs$amount data:

integer64
 [1] 517  200  391  186  262  1020 791  124  437  238  896  212  144  529  523  190

And I get something like this

> cut_data
 [1] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224] (0,224]
[15] (0,224] (0,224]
Levels: (0,224] (224,448] (448,672] (672,896] (896,1120]

What do I do wrong

EDIT:

result of dput() on rs$amount

structure(c(2.55431938899924e-321, 9.88131291682493e-322, 1.93179667523927e-321, 
9.18962101264719e-322, 1.29445199210407e-321, 5.03946958758071e-321, 
3.90805925860426e-321, 6.12641400843146e-322, 2.15906687232625e-321, 
1.17587623710217e-321, 4.42682818673757e-321, 1.04741916918344e-321, 
7.11454530011395e-322, 2.61360726650019e-321, 2.58396332774972e-321, 
9.38724727098368e-322), class = "integer64")

EDIT2:

Casting rs$amount as numeric helped with the issue

cut_data <- cut(as.numeric(rs$amount),points,dig.lab = 10)

Upvotes: 1

Views: 227

Answers (1)

r2evans
r2evans

Reputation: 161085

I think you have two alternatives: use cut(as.numeric(vec),...) or findInterval.

as.numeric

If you are not concerned about hitting the theoretical precision loss when converting to integer64 to numeric (it might be hard to find this happening), then you can convert to numeric:

cut(as.numeric(vec), points ,dig.lab = 10)
#  [1] (448,672]  (0,224]    (224,448]  (0,224]    (224,448]  (896,1120] (672,896]  (0,224]    (224,448]  (224,448]  (672,896]  (0,224]    (0,224]    (448,672]  (448,672]  (0,224]   
# Levels: (0,224] (224,448] (448,672] (672,896] (896,1120]

findInterval

table(cut(vec, points ,dig.lab = 10))
#    (0,224]  (224,448]  (448,672]  (672,896] (896,1120] 
#         16          0          0          0          0 
table(findInterval(vec, points))
# 1 2 3 4 5 
# 6 4 3 1 2

You can mock this to produce similarly-formatted factors manually:

labels <- sprintf("(%i,%i]", points[-length(points)], points[-1])
labels
# [1] "(0,224]"    "(224,448]"  "(448,672]"  "(672,896]"  "(896,1120]"
factor(labels[findInterval(vec, points)], labels = labels)
#  [1] (448,672]  (0,224]    (224,448]  (0,224]    (224,448]  (896,1120] (672,896]  (0,224]    (224,448]  (224,448]  (896,1120] (0,224]    (0,224]    (448,672]  (448,672]  (0,224]   
# Levels: (0,224] (224,448] (448,672] (672,896] (896,1120]

Data

vec <- structure(c(2.55431938899924e-321, 9.88131291682493e-322, 1.93179667523927e-321, 9.18962101264719e-322, 1.29445199210407e-321, 5.03946958758071e-321, 3.90805925860426e-321, 6.12641400843146e-322, 2.15906687232625e-321, 1.17587623710217e-321, 4.42682818673757e-321, 1.04741916918344e-321, 7.11454530011395e-322, 2.61360726650019e-321, 2.58396332774972e-321, 9.38724727098368e-322), class = "integer64")

Upvotes: 3

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