plot histogram from a vector by ggplot2 in r

Question

i want to plot a histogram by ggplot 2 from a vector. the dataset is rivers from datasets packages

rivers
  [1]  735  320  325  392  524  450 1459  135  465  600  330  336  280  315  870  906  202  329
 [19]  290 1000  600  505 1450  840 1243  890  350  407  286  280  525  720  390  250  327  230
 [37]  265  850  210  630  260  230  360  730  600  306  390  420  291  710  340  217  281  352
 [55]  259  250  470  680  570  350  300  560  900  625  332 2348 1171 3710 2315 2533  780  280
 [73]  410  460  260  255  431  350  760  618  338  981 1306  500  696  605  250  411 1054  735
 [91]  233  435  490  310  460  383  375 1270  545  445 1885  380  300  380  377  425  276  210
[109]  800  420  350  360  538 1100 1205  314  237  610  360  540 1038  424  310  300  444  301
[127]  268  620  215  652  900  525  246  360  529  500  720  270  430  671 1770

at first I tried these but didn't work:

> ggplot(rivers,aes(rivers))+geom_histogram()
Error: ggplot2 doesn't know how to deal with data of class numeric
> ggplot(rivers)+geom_histogram(aes(rivers))
Error: ggplot2 doesn't know how to deal with data of class numeric

then i found a similar question and figured out that i can achieve my goal by:

ggplot()+aes(rivers)+geom_histogram()
or
ggplot()+geom_histogram(aes(rivers))

i read through ggplot help documentation and have following questions:

• why i got error when i claim dataset in either ggplot() or geom_histogram(),like ggplot(data=rivers)? Help docs indicates a vector will be coerced to a data frame by default and a dataset must be specified. My assumption is that when a dataset is not specified, the function will search for global environment?
• why it works when i call aes(rivers) independently or in geom_histogram() but i got error when i put it in gglot(). why the location of aes(rivers) matters in this case?

  ggplot(aes(rivers))+geom_histogram()
  Error: ggplot2 doesn't know how to deal with data of class uneval
  

Answer 1

You can also use qplot() which is a shortcut for producing plots quickly. It is a function of ggplot2. The following code will solve your problem:

qplot(rivers, geom="histogram") 

Answer 2

If you don't provide argument name to ggplot explicitly, then value is assigned to the wrong ggplot argument (first in sequential order is data arg). Arguments order can be seen on the relevant function help page:

?ggplot 

ggplot(data = NULL, mapping = aes(), ..., environment = parent.frame())

Therefore when you provide aes() to ggplot without specifying argument name, it's like if you do the following:

ggplot(data = aes(rivers)) + geom_histogram()

since data argument don't allow this data type - you get an error. providing correct argument name solves the problem:

ggplot(mapping = aes(rivers)) + geom_histogram()

Answer 3

The reason for the error is rivers is a vector.

ggplot(aes(rivers))+
               geom_histogram()

Error: ggplot2 doesn't know how to deal with data of class uneval. Did you accidentally provide the results of aes() to the data argument?

Convert it to data.frame and then it would work

library(ggplot2)
library(dplyr)
data_frame(val = rivers) %>%
          ggplot(., aes(val)) + 
                geom_histogram()

data

set.seed(24)
rivers <- sample(700:1700, 150)