基于预先指定水平的 R 切割
R cut based on prespecified levels
我正在尝试削减我的数值,这样我也可以计算零的数量。不知道如何做到这一点。这些是我的目标。
1) I specifically get a count of number of zeros.
2) Option to cut the remaining non-zero values into many different
bins.
现在我在下面尝试了这个,但我无法得到任何零的数量。
c1 <- cut(df$Col1, breaks = seq(0, 1442, by = 53.25))
预期输出
(0] (0,53.2] (53.2,106] (106,160] (160,213] (213,266] (266,320] (320,373] (373,426] (426,479]
1652 1 6 1 34 6 1 1 8 2
(479,532] (532,586] (586,639] (639,692] (692,746] (746,799] (799,852] (852,905] (905,958]
0 0 4 1 0 0 1 0 0
(958,1.01e+03] (1.01e+03,1.06e+03] (1.06e+03,1.12e+03] (1.12e+03,1.17e+03] (1.17e+03,1.22e+03] (1.22e+03,1.28e+03] (1.28e+03,1.33e+03] (1.33e+03,1.38e+03] (1.38e+03,1.44e+03]
0 0 0 0 0 0 0 0 0
dput(df$Col1)
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205, 595, 0, 608, 601, 183, 381.56, 421, 1442, 408), label = "Col1", class = c("labelled",
"numeric"))
每个 bin 左侧 (0,53.2]
的 (
表示 "open-end",表示该边界 以上 的值。 (x
是你的 df$Col1
。)
看起来你想要 cut
的 table
,所以这是起点:
head(table(cut(x, breaks = seq(0, 1442, by = 53.25))))
# (0,53.2] (53.2,106] (106,160] (160,213] (213,266] (266,320]
# 1 6 1 34 6 1
两种选择。要么使用右关闭:
head(table(cut(x, breaks = seq(0, 1442, by = 53.25), right = FALSE)))
# [0,53.2) [53.2,106) [106,160) [160,213) [213,266) [266,320)
# 1652 7 1 32 8 1
(请注意,这会改变您的一些 bin 计数,如您在上面所见。)或者明确提供第一个 bin 的 "to the left" 内容:
head(table(cut(x, breaks = c(-Inf, seq(0, 1442, by = 53.25)))))
# (-Inf,0] (0,53.2] (53.2,106] (106,160] (160,213] (213,266]
# 1652 1 6 1 34 6
这会保留原始 bin 计数并确保您拥有所有零(以及,如果存在,任何负值)。
我正在尝试削减我的数值,这样我也可以计算零的数量。不知道如何做到这一点。这些是我的目标。
1) I specifically get a count of number of zeros.
2) Option to cut the remaining non-zero values into many different
bins.
现在我在下面尝试了这个,但我无法得到任何零的数量。
c1 <- cut(df$Col1, breaks = seq(0, 1442, by = 53.25))
预期输出
(0] (0,53.2] (53.2,106] (106,160] (160,213] (213,266] (266,320] (320,373] (373,426] (426,479]
1652 1 6 1 34 6 1 1 8 2
(479,532] (532,586] (586,639] (639,692] (692,746] (746,799] (799,852] (852,905] (905,958]
0 0 4 1 0 0 1 0 0
(958,1.01e+03] (1.01e+03,1.06e+03] (1.06e+03,1.12e+03] (1.12e+03,1.17e+03] (1.17e+03,1.22e+03] (1.22e+03,1.28e+03] (1.28e+03,1.33e+03] (1.33e+03,1.38e+03] (1.38e+03,1.44e+03]
0 0 0 0 0 0 0 0 0
dput(df$Col1)
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 198, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 204, 0, 0, 0, 207, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 209, 0, 0, 161.19, 0, 0, 106, 0, 0, 0, 0, 0, 0, 0,
0, 100, 0, 0, 0, 0, 0, 0, 0, 200, 0, 0, 0, 195, 0, 0, 0, 0, 398,
0, 0, 0, 0, 0, 103, 0, 0, 0, 0, 0, 0, 204, 0, 89.37, 0, 0, 0,
0, 0, 0, 194, 0, 0, 0, 0, 212, 0, 0, 0, 0, 212, 211, 0, 402,
219, 0, 0, 244, 194, 0, 183.75, 0, 0, 0, 0, 0, 0, 0, 104, 197,
0, 0, 53.25, 0, 0, 0, 0, 0, 0, 0, 103, 0, 0, 0, 0, 0, 0, 383,
314, 202, 0, 0, 0, 0, 204, 227, 0, 205, 211, 670, 230.39, 0,
0, 110, 801, 595, 0, 0, 0, 438, 0, 397, 203, 209, 0, 209, 0,
258, 0, 0, 213, 0, 201, 174.84, 213, 0, 407, 208, 218, 365.7,
205, 595, 0, 608, 601, 183, 381.56, 421, 1442, 408), label = "Col1", class = c("labelled",
"numeric"))
每个 bin 左侧 (0,53.2]
的 (
表示 "open-end",表示该边界 以上 的值。 (x
是你的 df$Col1
。)
看起来你想要 cut
的 table
,所以这是起点:
head(table(cut(x, breaks = seq(0, 1442, by = 53.25))))
# (0,53.2] (53.2,106] (106,160] (160,213] (213,266] (266,320]
# 1 6 1 34 6 1
两种选择。要么使用右关闭:
head(table(cut(x, breaks = seq(0, 1442, by = 53.25), right = FALSE)))
# [0,53.2) [53.2,106) [106,160) [160,213) [213,266) [266,320)
# 1652 7 1 32 8 1
(请注意,这会改变您的一些 bin 计数,如您在上面所见。)或者明确提供第一个 bin 的 "to the left" 内容:
head(table(cut(x, breaks = c(-Inf, seq(0, 1442, by = 53.25)))))
# (-Inf,0] (0,53.2] (53.2,106] (106,160] (160,213] (213,266]
# 1652 1 6 1 34 6
这会保留原始 bin 计数并确保您拥有所有零(以及,如果存在,任何负值)。