计算一个列中另一列分组的类别值
[英]Counting categorical values in one column grouped by another column
问题描述
R新手在这里，所以请原谅我的无知。 我的数据如下所示：
JOB_ROLE EXP_IT_NETW
1 Software engineering-related (developer, tester, project manager, architecture) 5<10
3 See below None
4 Student <1
5 Software engineering-related (developer, tester, project manager, architecture) 1<5
6 Blogger 10+
我想对第2列中每个值的实例进行计数，并按第1列对计数进行分组，以便结果看起来像这样：
JOB_ROLE None <1 1<5 5<10 10+
Software engineer 3 5 10 15 3
Student 10 7 5 1 0
...
有关如何执行此操作的任何想法？ 我的dput输出如下。 提前致谢！
structure(list(JOB_ROLE = c("Software engineering-related (developer, tester, project manager, architecture)",
"See below", "Student", "Software engineering-related (developer, tester, project manager, architecture)",
"Blogger", "Systems Support", "Student", "IT/Network Administrator",
"Software engineering-related (developer, tester, project manager, architecture)",
"Student", "Student", "Software engineering-related (developer, tester, project manager, architecture)",
"IT hobbyist", "Student", "Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"IT Manager", "Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"IT/Network Administrator", "IT/Network Administrator", "Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"Student", "Software engineering-related (developer, tester, project manager, architecture)",
"Researcher in CompSci or related field", "Researcher in CompSci or related field",
"IT/Network Administrator", "Student", "Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"Education", "Software engineering-related (developer, tester, project manager, architecture)",
"Software engineering-related (developer, tester, project manager, architecture)",
"IT/Network Administrator", "Software engineering-related (developer, tester, project manager, architecture)",
"IT/Network Administrator", "Student", "IT/Network Administrator",
"Software engineering-related (developer, tester, project manager, architecture)",
"Student", "IT/Network Administrator", "just a layperson who has used computers for over 30 years",
"IT/Network Administrator", "Unemployed", "Student", "IT/Network Administrator"
), EXP_IT_NETW = c("5<10", "None", "<1", "1<5", "10+", "None",
"1<5", "10+", "<1", "None", "1<5", "1<5", "None", "None", "10+",
"None", "1<5", "10+", "None", "1<5", "None", "1<5", "10+", "1<5",
"1<5", "1<5", "None", "None", "1<5", "5<10", "None", "5<10",
"<1", "None", "1<5", "None", "1<5", "1<5", "10+", "1<5", "10+",
"None", "1<5", "5<10", "None", "1<5", "None", "1<5", "None",
"None", "10+")), .Names = c("JOB_ROLE", "EXP_IT_NETW"), class = "data.frame", row.names = c(1L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 16L, 17L, 18L,
19L, 20L, 21L, 22L, 23L, 25L, 26L, 27L, 28L, 29L, 30L, 32L, 33L,
34L, 35L, 36L, 37L, 39L, 40L, 41L, 42L, 43L, 44L, 47L, 48L, 49L,
50L, 51L, 52L, 53L, 55L, 56L, 57L, 59L, 61L, 62L))
2 个解决方案
解决方案1
5 已采纳 2013-03-27 19:57:16
使用table ：
> table(d)
EXP_IT_NETW
JOB_ROLE <1 1<5 10+ 5<10 None
Blogger 0 0 1 0 0
Education 0 0 0 0 1
IT hobbyist 0 0 0 0 1
IT Manager 0 1 0 0 0
IT/Network Administrator 0 4 5 1 0
just a layperson who has used computers for over 30 years 0 0 0 0 1
Researcher in CompSci or related field 0 1 0 0 1
See below 0 0 0 0 1
Software engineering-related (developer, tester, project manager, architecture) 2 9 2 3 5
Student 1 3 0 0 6
Systems Support 0 0 0 0 1
Unemployed 0 0 0 0 1
解决方案2
3 2013-03-27 19:55:27
我也会使用data.table但是要获得您期望的相同格式有些不同。
require(data.table)
dt <- data.table(df) # here, I assume df is your data.frame
setkey(dt, "JOB_ROLE") # setkey for fast access/grouping
dt[, {tt <- table(factor(EXP_IT_NETW,
levels=factor(unique(dt$EXP_IT_NETW))));
setattr(as.list(tt), 'names', names(tt))
}, by = key(dt)]
我得到这个：
# JOB_ROLE None 10+ 1<5 5<10 <1
# 1: >30_years_experience 1 0 0 0 0
# 2: Blogger 0 1 0 0 0
# 3: Education 1 0 0 0 0
# 4: IT Manager 0 0 1 0 0
# 5: IT hobbyist 1 0 0 0 0
# 6: IT/Network Administrator 0 5 4 1 0
# 7: Researcher in CompSci or related field 1 0 1 0 0
# 8: See below 1 0 0 0 0
# 9: Software_enginnering 5 2 9 3 2
# 10: Student 6 0 3 0 1
# 11: Systems Support 1 0 0 0 0
# 12: Unemployed 1 0 0 0 0
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