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How to fill geom_polygon with different colors above and below y = 0?

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Considering the following polygon plot:

ggplot(df, aes(x=year,y=afw)) +
  geom_polygon() +
  scale_x_continuous("", expand=c(0,0), breaks=seq(1910,2010,10)) +
  theme_bw()

enter image description here

However, i want to fill this with two different colors. For example red for the black areas above 0 and blue for the black areas below 0. Unfortunately, using fill=col doesn't fill the correct areas.

I tried the following code (I added the geom_line in order to illustrate where the border of the fill should be):

ggplot(df, aes(x=year,y=afw)) +
  geom_line() +
  geom_polygon(aes(fill=col), alpha=0.5) +
  scale_x_continuous("", expand=c(0,0), breaks=seq(1910,2010,10)) +
  theme_bw()

which gives: enter image description here

As you can see, it's filling a lot more than it's supposed to do. How can I solve this?

The data:

df <- structure(list(year = c(1901, 1901, 1901, 1902, 1903, 1904, 1905, 1906, 1907, 1908, 1909, 1910, 1911, 1912, 1913, 1914, 1915, 1916, 1917, 1918, 1919, 1920, 1921, 1922, 1923, 1924, 1925, 1926, 1927, 1928, 1929, 1930, 1931, 1932, 1933, 1934, 1935, 1936, 1937, 1938, 1939, 1940, 1941, 1942, 1943, 1944, 1945, 1946, 1947, 1948, 1949, 1950, 1951, 1952, 1953, 1954, 1955, 1956, 1957, 1958, 1959, 1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2013, 2013), afw = c(0, 0, -0.246246074793035, -2.39463317156723, -2.39785897801884, 0.840850699400514, -0.843020268341422, -3.02043962318013, -0.033342848986583, -2.04947188124465, -0.00431059092206709, 2.49568940907793, 1.96988295746503, 2.26665715101342, 0.986011989723095, 1.79568940907793, 2.06665715101342, -0.601084784470454, -3.21076220382529, 2.65052811875535, 0.46988295746503, -1.09140736511562, 0.0505281187553526, 1.41827005423922, -2.80108478447045, 0.611818441335997, -1.83011704253497, -0.30753639737368, -4.43011704253497, -0.897858978018841, 1.98601198972309, -0.965600913502712, 0.0795603768198685, 0.308592634884385, -5.33011704253497, 4.00214102198116, -0.594633171567228, 0.0698829574650297, -1.60753639737368, -2.81398801027691, -2.21398801027691, -2.4365686554382, 1.53439908649729, 1.06665715101342, -1.87205252640594, -0.688181558664002, 0.0569797316585783, -3.51398801027691, 0.979560376819868, 0.289237796174707, 1.24085069940051, -4.39140736511562, 1.13117328004567, -1.72689123608336, 2.20214102198116, 2.27310876391664, 1.46665715101342, 2.18278618327148, -0.23011704253497, 1.50536682843277, 1.17633457036826, -0.0785041393091639, -1.54947188124465, -3.85269768769626, -4.31398801027691, -0.80753639737368, 1.27956037681987, 1.2376248929489, 0.195689409077933, -3.38172994576078, -4.88172994576078, -0.675278332857551, 2.25375392520697, 0.0924636026263199, -0.446246074793035, 4.06988295746503, 0.350528118755352, -1.48172994576078, 1.81504424778761, -1.42689123608336, 2.22472166714245, 0.376334570368256, -3.88495575221239, 0.211818441335998, 0.586011989723094, 1.14407650585213, 2.55697973165858, 1.92794747359406, 1.20214102198116, 3.83439908649729, 1.64407650585213, 0.986011989723095, 0.753753925206965, 0.508592634884385, 1.911818441336, 2.11504424778761, -4.06560091350271, -2.58495575221239, 1.80859263488438, 1.37956037681987, 1.58923779617471, 1.88601198972309, -0.323665429631744, -0.291407365115615, 0.818270054239223, 0.0569797316585783, 0.795689409077933, 3.32472166714245, 0.595689409077933, -0.733342848986583, -0.955923494147874, -4.32689123608336, 3.29891521552955, 1.85697973165858, 2.74407650585213, 0, 0), col = structure(c(1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L), .Label = c("B", "A"), class = "factor")), .Names = c("year", "afw", "col"), class = c("tbl_df", "data.frame"), row.names = c(NA, -117L))

Note: as you can see in the data, there are 3 rows for both 1901 and 2013. I did this because I wanted to get the fill right. Although the black fill is correct, I seem not to get a working solution with colors.

The original dataset:

orig <- structure(list(year = c(1901, 1902, 1903, 1904, 1905, 1906, 1907, 1908, 1909, 1910, 1911, 1912, 1913, 1914, 1915, 1916, 1917, 1918, 1919, 1920, 1921, 1922, 1923, 1924, 1925, 1926, 1927, 1928, 1929, 1930, 1931, 1932, 1933, 1934, 1935, 1936, 1937, 1938, 1939, 1940, 1941, 1942, 1943, 1944, 1945, 1946, 1947, 1948, 1949, 1950, 1951, 1952, 1953, 1954, 1955, 1956, 1957, 1958, 1959, 1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013), afw = c(-0.246246074793035, -2.39463317156723, -2.39785897801884, 0.840850699400514, -0.843020268341422, -3.02043962318013, -0.033342848986583, -2.04947188124465, -0.00431059092206709, 2.49568940907793, 1.96988295746503, 2.26665715101342, 0.986011989723095, 1.79568940907793, 2.06665715101342, -0.601084784470454, -3.21076220382529, 2.65052811875535, 0.46988295746503, -1.09140736511562, 0.0505281187553526, 1.41827005423922, -2.80108478447045, 0.611818441335997, -1.83011704253497, -0.30753639737368, -4.43011704253497, -0.897858978018841, 1.98601198972309, -0.965600913502712, 0.0795603768198685, 0.308592634884385, -5.33011704253497, 4.00214102198116, -0.594633171567228, 0.0698829574650297, -1.60753639737368, -2.81398801027691, -2.21398801027691, -2.4365686554382, 1.53439908649729, 1.06665715101342, -1.87205252640594, -0.688181558664002, 0.0569797316585783, -3.51398801027691, 0.979560376819868, 0.289237796174707, 1.24085069940051, -4.39140736511562, 1.13117328004567, -1.72689123608336, 2.20214102198116, 2.27310876391664, 1.46665715101342, 2.18278618327148, -0.23011704253497, 1.50536682843277, 1.17633457036826, -0.0785041393091639, -1.54947188124465, -3.85269768769626, -4.31398801027691, -0.80753639737368, 1.27956037681987, 1.2376248929489, 0.195689409077933, -3.38172994576078, -4.88172994576078, -0.675278332857551, 2.25375392520697, 0.0924636026263199, -0.446246074793035, 4.06988295746503, 0.350528118755352, -1.48172994576078, 1.81504424778761, -1.42689123608336, 2.22472166714245, 0.376334570368256, -3.88495575221239, 0.211818441335998, 0.586011989723094, 1.14407650585213, 2.55697973165858, 1.92794747359406, 1.20214102198116, 3.83439908649729, 1.64407650585213, 0.986011989723095, 0.753753925206965, 0.508592634884385, 1.911818441336, 2.11504424778761, -4.06560091350271, -2.58495575221239, 1.80859263488438, 1.37956037681987, 1.58923779617471, 1.88601198972309, -0.323665429631744, -0.291407365115615, 0.818270054239223, 0.0569797316585783, 0.795689409077933, 3.32472166714245, 0.595689409077933, -0.733342848986583, -0.955923494147874, -4.32689123608336, 3.29891521552955, 1.85697973165858, 2.74407650585213)), .Names = c("year", "afw"), class = c("tbl_df", "data.frame"), row.names = c(NA, -113L))
like image 887
Jaap Avatar asked Nov 25 '14 20:11

Jaap


2 Answers

Get the indices where the y value of two consecutive time steps have different sign. Use linear interpolation between these points to generate new x values where y is zero.

First, a smaller example to make it easier to get a feeling for the linear interpolation and which points are added to the original data:

# original data
d <- data.frame(x = 1:6,
                y = c(-1, 2, 1, 2, -1, 1))

# coerce to data.table
library(data.table)
setDT(d)

# make sure data is ordered by x
setorder(d, x)

# add a grouping variable
# only to keep track of original and interpolated points in this example
d[ , g := "orig"]

# interpolation
d2 = d[ , {
  ix = .I[c(FALSE, abs(diff(sign(d$y))) == 2)]
  if(length(ix)){
    pred_x = sapply(ix, function(i) approx(x = y[c(i-1, i)], y = x[c(i-1, i)], xout = 0)$y)
    rbindlist(.(.SD, data.table(x = pred_x, y = 0, g = "new")))} else .SD
}]

d2   
#           x  y  grp
# 1  1.000000 -1 orig
# 2  2.000000  2 orig
# 3  3.000000  1 orig
# 4  4.000000  2 orig
# 5  5.000000 -1 orig
# 6  6.000000  1 orig
# 13 1.333333  0  new
# 11 4.666667  0  new
# 12 5.500000  0  new

Plot with original and new points differentiated by color:

ggplot(data = d2, aes(x = x, y = y)) +
  geom_area(data = d2[y <= 0], fill = "red", alpha = 0.2) +
  geom_area(data = d2[y >= 0], fill = "blue", alpha = 0.2) +
  geom_point(aes(color = g), size = 4) +
  scale_color_manual(values = c("red", "black")) +
  theme_bw()

enter image description here


Apply on OP's data:

d = as.data.table(orig)
# setorder(d, year)

d2 = d[ , {
  ix = .I[c(FALSE, abs(diff(sign(d$afw))) == 2)]
  if(length(ix)){
    pred_yr = sapply(ix, function(i) approx(afw[c(i-1, i)], year[c(i-1, i)], xout = 0)$y)
    rbindlist(.(.SD, data.table(year = pred_yr, afw = 0)))} else .SD}]

ggplot(data = d2, aes(x = year, y = afw)) +
  geom_area(data = d2[afw <= 0], fill = "red") +
  geom_area(data = d2[afw >= 0], fill = "blue") +
  theme_bw()

enter image description here


In reply to @Jason Whythe's comment, the method above can be modified to account for grouped data. The interpolation is made within each group, and the plot is facetted by group:

# data grouped by 'id' 
d = data.table(
  id = rep(c("a", "b", "c"), c(6, 5, 4)),
  x = as.numeric(c(1:6, 1:5, 1:4)),
  y = c(-1, 2, 1, 2, -1, 1,
        0, -2, 0, -1, -2, 
        2, 1, -1, 1.5))

# again, this variable is just added for illustration 
d[ , g := "orig"]

d2 = d[ , {
  ix = .I[c(FALSE, abs(diff(sign(.SD$y))) == 2)]
  if(length(ix)){
    pred_x = sapply(ix, function(i) approx(x = d$y[c(i-1, i)], y = d$x[c(i-1, i)], xout = 0)$y)
    rbindlist(.(.SD, data.table(x = pred_x, y = 0, g = "new")))} else .SD
}, by = id]

ggplot(data = d2, aes(x = x, y = y)) +
  facet_wrap(~ id) +
  geom_area(data = d2[y <= 0], fill = "red", alpha = 0.2) +
  geom_area(data = d2[y >= 0], fill = "blue", alpha = 0.2) +
  geom_point(aes(color = g), size = 4) +
  scale_color_manual(values = c("red", "black")) +
  theme_bw()

enter image description here


For an alternative base solution adapted from @kohske's answer here (credits to him), see previous edits.

like image 156
Henrik Avatar answered Nov 10 '22 11:11

Henrik


So this is not perfect and I'm interested to see what others come up with...

The reason for the "multiple" colored areas is that a single polygon is bounded by the data points and the data points are not actually zero.

To solve this, we can interpolate using approx(). For a perfect solution, you would need to determine exactly where the line crosses zero.

interp <- approx(orig$year, orig$afw, n=10000)

orig2 <- data.frame(year=interp$x, afw=interp$y)
orig2$col[orig2$afw >= 0] <- "pos"
orig2$col[orig2$afw < 0] <- "neg"

ggplot(orig2, aes(x=year, y=afw)) +
  geom_area(aes(fill=col)) +
  geom_line() +
  geom_hline(yintercept=0)

Solution

However, you will see this still has issues when you zoom:

Zoomed


To elaborate on my statement above (and further illustrate the original "problem/issue"), consider what happens when you plot each of the original positive and negative datasets separately:

p1 <- ggplot(subset(orig, col == "neg"), aes(x = year, y = afw)) +
  geom_area(aes(fill=col)) +
  scale_fill_manual(values = c("#FF3030", "#00CC66"))

p2 <- ggplot(subset(orig, col == "pos"), aes(x = year, y = afw)) +
  geom_area(aes(fill=col)) +
  scale_fill_manual(values = c("#00CC66", "#FF3030"))

library(gridExtra)

grid.arrange(p2, p1)

Multiple Plots


Of course, you could always solve this by utilizing a different type of visualization:

ggplot(data = orig, aes(x = year, y = afw)) +
  geom_bar(stat = "identity", aes(fill=col), colour = "white")

Alternate Solution

like image 38
JasonAizkalns Avatar answered Nov 10 '22 12:11

JasonAizkalns