Question

编辑感谢@ user5249203指出地理编码最好用ggmaps＆＃39;地理编码呼叫。但请注意NA。

我正在与R中的apply家庭挣扎。

我使用的是function，它接收一个字符串并返回经度和纬度

> gGeoCode("Philadelphia, PA") [1] 39.95258 -75.16522

我有一个简单的数据框，其中包含所有52个状态的名称：

dput(state_lat_long)
structure(
  list(State = structure(
    c(
      32L, 28L, 43L, 5L, 23L, 34L,
      30L, 13L, 14L, 38L, 22L, 25L, 15L, 20L, 24L, 40L, 46L, 21L, 9L,
      18L, 48L, 10L, 7L, 4L, 3L, 31L, 35L, 37L, 49L, 44L, 12L, 6L,
      17L, 36L, 11L, 39L, 42L, 8L, 47L, 33L, 16L, 1L, 29L, 27L, 26L,
      19L, 41L, 50L, 2L, 45L
    ), .Label = c(
      "alabama", "alaska", "arizona",
      "arkansas", "california", "colorado", "connecticut", "delaware",
      "florida", "georgia", "hawaii", "idaho", "illinois", "indiana",
      "iowa", "kansas", "kentucky", "louisiana", "maine", "maryland",
      "massachusetts", "michigan", "minnesota", "mississippi", "missouri",
      "montana", "nebraska", "nevada", "new hampshire", "new jersey",
      "new mexico", "new york", "north carolina", "north dakota", "ohio",
      "oklahoma", "oregon", "pennsylvania", "rhode island", "south carolina",
      "south dakota", "tennessee", "texas", "utah", "vermont", "virginia",
      "washington", "west virginia", "wisconsin", "wyoming"
    ), class = "factor"
  )), .Names = "State", row.names = c(NA,-50L), class = "data.frame"
)

要练习apply技能，我只想将gGeoCode应用于state_lat_long数据框唯一列中的每个单元格。

不会简单得多。

然后这有什么问题？

> View(apply(state_lat_long, function(x) gGeoCode(x)))

当我跑步时，我得到：

Error in View : argument "FUN" is missing, with no default

我不明白，因为FUN并没有丢失。

所以，让我们试试sapply。它应该很简单，对吧？

但是这有什么问题？

View(sapply(state_lat_long$State, function(x) gGeoCode(x)))

当我运行这个时，我得到2行50列，包含NA s。我无法理解它。

接下来，我试过

View(apply(state_lat_long, 2, function(x) gGeoCode(x)))

我得到了

     State
  40.71278
 -74.00594

同样，这没有任何意义！

我做错了什么？谢谢。

Answer 1

这是您的数据框架的方式吗？

df = data.frame(State = c(
    32L, 28L, 43L, 5L, 23L, 34L,
    30L, 13L, 14L, 38L, 22L, 25L, 15L, 20L, 24L, 40L, 46L, 21L, 9L,
    18L, 48L, 10L, 7L, 4L, 3L, 31L, 35L, 37L, 49L, 44L, 12L, 6L,
    17L, 36L, 11L, 39L, 42L, 8L, 47L, 33L, 16L, 1L, 29L, 27L, 26L,
    19L, 41L, 50L, 2L, 45L
  ), Label = c(
    "alabama", "alaska", "arizona",
    "arkansas", "california", "colorado", "connecticut", "delaware",
    "florida", "georgia", "hawaii", "idaho", "illinois", "indiana",
    "iowa", "kansas", "kentucky", "louisiana", "maine", "maryland",
    "massachusetts", "michigan", "minnesota", "mississippi", "missouri",
    "montana", "nebraska", "nevada", "new hampshire", "new jersey",
    "new mexico", "new york", "north carolina", "north dakota", "ohio",
    "oklahoma", "oregon", "pennsylvania", "rhode island", "south carolina",
    "south dakota", "tennessee", "texas", "utah", "vermont", "virginia",
    "washington", "west virginia", "wisconsin", "wyoming"
  ))

head(df)
  State      Label
1    32    alabama
2    28     alaska
3    43    arizona
4     5   arkansas
5    23 california
6    34   colorado

apply(df, 1, function(x) gGeoCode(x))

或者，

mapply(FUN = gGeoCode, df$Label, SIMPLIFY = T)

注意：某些州仍然会抛出NA。重新运行代码将获取缺少的坐标。但是，如果我们知道您的输入格式/数据帧结构，我希望它能更有效地工作。此外，确保您传递的参数是gGeoCode期望的参数非常重要。

Answer 2

我意识到这个问题主要是关于*apply，但是，如果您只是在地理编码之后，更容易的选择是使用向量化函数，例如ggmap::geocode

state_lat_long <- structure(
    list(State = structure(
    c(
      32L, 28L, 43L, 5L, 23L, 34L,
      30L, 13L, 14L, 38L, 22L, 25L, 15L, 20L, 24L, 40L, 46L, 21L, 9L,
      18L, 48L, 10L, 7L, 4L, 3L, 31L, 35L, 37L, 49L, 44L, 12L, 6L,
      17L, 36L, 11L, 39L, 42L, 8L, 47L, 33L, 16L, 1L, 29L, 27L, 26L,
      19L, 41L, 50L, 2L, 45L
    ), .Label = c(
      "alabama", "alaska", "arizona",
      "arkansas", "california", "colorado", "connecticut", "delaware",
      "florida", "georgia", "hawaii", "idaho", "illinois", "indiana",
      "iowa", "kansas", "kentucky", "louisiana", "maine", "maryland",
      "massachusetts", "michigan", "minnesota", "mississippi", "missouri",
      "montana", "nebraska", "nevada", "new hampshire", "new jersey",
      "new mexico", "new york", "north carolina", "north dakota", "ohio",
      "oklahoma", "oregon", "pennsylvania", "rhode island", "south carolina",
      "south dakota", "tennessee", "texas", "utah", "vermont", "virginia",
      "washington", "west virginia", "wisconsin", "wyoming"
    ), class = "factor"
  )), .Names = "State", row.names = c(NA,-50L), class = "data.frame"
)

library(ggmap)

## to make sure we're using the correct geocode function I call it with 'ggmap::geocode'
ggmap::geocode(as.character(state_lat_long$State))
...
#           lon      lat
# 1   -74.00594 40.71278
# 2  -116.41939 38.80261
# 3   -99.90181 31.96860
# 4  -119.41793 36.77826
# 5   -94.68590 46.72955
# 6  -101.00201 47.55149

将函数应用于R中数据帧的列中的每个单元格

2 个答案: