I have a big table of total 969 rows and I need to find the difference between every two rows, e.g. row1 and row2, row2 and row3, row3 and row4 etc. How can I do that? I was told to do it by the command diff() but I have no idea where to start.
我有一个总共969行的大表，我需要找到每两行之间的差异，例如row1和row2，row2和row3，row3和row4等。我该怎么做？我被告知通过命令diff（）来做它，但我不知道从哪里开始。
3 个解决方案
#1
32
Here's an example of how to use diff() on the built-in mtcars data.frame. You have to select a column to perform the diff over:
这是一个如何在内置mtcars data.frame上使用diff（）的示例。您必须选择一列来执行差异：
> mtcars
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
> diff(mtcars$qsec)
[1] 0.56 1.59 0.83 -2.42 3.20 -4.38 4.16 2.90 -4.60 0.60 -1.50 0.20
[13] 0.40 -0.02 -0.16 -0.40 2.05 -0.95 1.38 0.11 -3.14 0.43 -1.89 1.64
[25] 1.85 -2.20 0.20 -2.40 1.00 -0.90 4.00
#2
12
You could simply subtract a data.frame consisting of rows 1:(n-1) of the original data.frame from a second one consisting of rows 2:n. (Here n is the number of rows in the original data.frame):
您可以简单地从由行2：n组成的第二行中删除包含原始data.frame的行1：（n-1）的data.frame。 （这里n是原始data.frame中的行数）：
# Example data
df <- data.frame(a=1:4, b=4:1, c=11:14, d=c(2,4,10,0))
# a b c d
# 1 1 4 11 2
# 2 2 3 12 4
# 3 3 2 13 10
# 4 4 1 14 0
# Calculate the differences
diff_df <- df[-1,] - df[-nrow(df),]
diff_df
# a b c d
# 2 1 -1 1 2
# 3 1 -1 1 6
# 4 1 -1 1 -10
You can then rename the rows as you see fit using something like:
然后，您可以使用以下内容重命名行：
row.names(diff_df) <- paste("d", seq_len(nrow(diff_df)), sep="")
diff_df
# a b c d
# d1 1 -1 1 2
# d2 1 -1 1 6
# d3 1 -1 1 -10
#3
3
This would have be clearer with sample data. Let's assume you mean "numerical difference", and your data can be represented as a matrix, this would do.
样本数据会更清楚。假设你的意思是“数字差异”，你的数据可以表示为矩阵，这样就可以了。
set.seed(4871)
m = matrix(sample(1:5,50,TRUE),nrow=10,ncol=5)
m
t(apply(m,1,diff))
#1
32
Here's an example of how to use diff() on the built-in mtcars data.frame. You have to select a column to perform the diff over:
这是一个如何在内置mtcars data.frame上使用diff（）的示例。您必须选择一列来执行差异：
> mtcars
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
> diff(mtcars$qsec)
[1] 0.56 1.59 0.83 -2.42 3.20 -4.38 4.16 2.90 -4.60 0.60 -1.50 0.20
[13] 0.40 -0.02 -0.16 -0.40 2.05 -0.95 1.38 0.11 -3.14 0.43 -1.89 1.64
[25] 1.85 -2.20 0.20 -2.40 1.00 -0.90 4.00
#2
12
You could simply subtract a data.frame consisting of rows 1:(n-1) of the original data.frame from a second one consisting of rows 2:n. (Here n is the number of rows in the original data.frame):
您可以简单地从由行2：n组成的第二行中删除包含原始data.frame的行1：（n-1）的data.frame。 （这里n是原始data.frame中的行数）：
# Example data
df <- data.frame(a=1:4, b=4:1, c=11:14, d=c(2,4,10,0))
# a b c d
# 1 1 4 11 2
# 2 2 3 12 4
# 3 3 2 13 10
# 4 4 1 14 0
# Calculate the differences
diff_df <- df[-1,] - df[-nrow(df),]
diff_df
# a b c d
# 2 1 -1 1 2
# 3 1 -1 1 6
# 4 1 -1 1 -10
You can then rename the rows as you see fit using something like:
然后，您可以使用以下内容重命名行：
row.names(diff_df) <- paste("d", seq_len(nrow(diff_df)), sep="")
diff_df
# a b c d
# d1 1 -1 1 2
# d2 1 -1 1 6
# d3 1 -1 1 -10
#3
3
This would have be clearer with sample data. Let's assume you mean "numerical difference", and your data can be represented as a matrix, this would do.
样本数据会更清楚。假设你的意思是“数字差异”，你的数据可以表示为矩阵，这样就可以了。
set.seed(4871)
m = matrix(sample(1:5,50,TRUE),nrow=10,ncol=5)
m
t(apply(m,1,diff))