我有一个数据集，同一ID可能具有同一变量的多个记录。其中一些具有部分NA，其中一些具有全部NA。
我想基于相同的ID对变量求和，并希望该变量中具有所有NA的人获得一个NA，而该变量中具有部分NA的人获得一个总和（在这种情况下，将NA处理为0）。有什么办法吗？
在对变量求和时，我尝试了na.rm = T，并且所有NA都变为0，这不是我想要的。
Dataset:
ID V1
5 120
5 300
5 NA
8 NA
8 NA
8 NA
Want this:
ID V1
5 420
8 NA
I did this and all NA became 0:
df <- df %>% group_by(ID) %>% transmute(V1 = sum(V1, na.rm = T))
答案 0 :(得分：2)
大多数方法都会删除import open3d as o3d
import numpy as np
import copy
def draw_registration_result(source, target, transformation):
source_temp = copy.deepcopy(source)
target_temp = copy.deepcopy(target)
source_temp.paint_uniform_color([1, 0.706, 0])
target_temp.paint_uniform_color([0, 0.651, 0.929])
source_temp.transform(transformation)
o3d.visualization.draw_geometries([source_temp, target_temp])
if __name__ == "__main__":
source = o3d.io.read_point_cloud("C:/Users/Kathan/Desktop/Biogen/NewScans/2/0deg.pcd")
target = o3d.io.read_point_cloud("C:/Users/Kathan/Desktop/Biogen/NewScans/2/180deg.pcd")
threshold = 0.05
trans_init = np.asarray([[0.0, 0.0, 3.0, 0.0], [1.0, 0.0, 0.0, 0.0], [0.0, 1.0, 0.0, 0.0], [0.0, 0.0, 0.0, 1.0]])
#trans_init = np.asarray([[0.862, 0.011, -0.507, 0.5],[-0.139, 0.967, -0.215, 0.7],[0.487, 0.255, 0.835, -1.4], [0.0, 0.0, 0.0, 1.0]])
draw_registration_result(source, target, trans_init)
print("Initial alignment")
evaluation = o3d.registration.evaluate_registration(source, target,
threshold, trans_init)
print(evaluation)
print("Apply point-to-point ICP")
reg_p2p = o3d.registration.registration_icp(
source, target, threshold, trans_init,
o3d.registration.TransformationEstimationPointToPoint())
print(reg_p2p)
print("Transformation is:")
print(reg_p2p.transformation)
print("")
draw_registration_result(source, target, reg_p2p.transformation)
print("Apply point-to-plane ICP")
reg_p2l = o3d.registration.registration_icp(
source, target, threshold, trans_init,
o3d.registration.TransformationEstimationPointToPlane())
print(reg_p2l)
print("Transformation is:")
print(reg_p2l.transformation)
print("")
draw_registration_result(source, target, reg_p2l.transformation)
组或将其设为0。也许我们可以使用自定义条件
NA，且基数为R library(dplyr)
df %>%
group_by(ID) %>%
summarise(V1 = if (all(is.na(V1))) NA else sum(V1, na.rm = TRUE))
# A tibble: 2 x 2
# ID V1
# <int> <int>
#1 5 420
#2 8 NA
aggregate答案 1 :(得分：0)
我们可以使用Math.random中的sum_，如果所有元素都是hablar，它将自动返回NA。使用NA语法，它将是
data.table或与library(data.table)
library(hablar)
setDT(df)[, .(V1 = sum_(V1)), .(ID)]
# ID V1
#1: 5 420
#2: 8 NA
dplyr或者使用library(dplyr)
df %>%
group_by(ID) %>%
summarise(V1 = sum_(V1))
# A tibble: 2 x 2
# ID V1
# <int> <int>
#1 5 420
#2 8 NA
而不使用任何sum
if/else或使用df %>%
group_by(ID) %>%
summarise(V1 = sum(V1, na.rm = TRUE) * NA^ all(is.na(V1)))
# A tibble: 2 x 2
# ID V1
# <int> <dbl>
#1 5 420
#2 8 NA
base R或与out <- rowsum(df$V1, df$ID, na.rm = TRUE)
(NA^!out) * out
# [,1]
#5 420
#8 NA
by注意：所有代码都很紧凑
by(df$V1, df$ID, FUN = sum_)