Infil NaN for missing years in time series

1 次查看（过去 30 天）
Poulomi Ganguli
Poulomi Ganguli 2019-8-9
评论： Neuropragmatist 2019-8-9
Hello:
I have two matrices, A and B. Matrix B has some years missing in it. I want to concatenate horizontally two matrices, A and B to a new matrix C, which can be filled with NaN for the missing years.
A =
1981 0.79 1.56 0.90 1.15
1982 0.62 0.83 0.84 0.74
1983 0.81 0.71 0.71 0.70
1984 1.06 0.74 0.61 0.76
1985 1.23 0.86 0.67 0.61
1986 1.32 0.56 1.11 0.76
1987 0.75 1.06 0.56 1.15
1988 1.76 1.09 0.88 0.67
1989 0.90 0.77 0.94 0.77
1990 0.52 0.52 1.15 0.88
and B =
1981 1.0617
1982 1.0682
1985 1.0149
1986 0.6607
1987 0.5642
1988 0.6194
1989 0.6693
1990 0.6966
Desired output, C =
1981 1.0617 0.79 1.56 0.90 1.15
1982 1.0682 0.62 0.83 0.84 0.74
1983 NaN 0.81 0.71 0.71 0.70
1984 NaN 1.06 0.74 0.61 0.76
1985 1.0149 1.23 0.86 0.67 0.61
1986 0.6607 1.32 0.56 1.11 0.76
1987 0.5642 0.75 1.06 0.56 1.15
1988 0.6194 1.76 1.09 0.88 0.67
1989 0.6693 0.90 0.77 0.94 0.77
1990 0.6966 0.52 0.52 1.15 0.88

请先登录，再回答此问题。

采纳的回答

Neuropragmatist
Neuropragmatist 2019-8-9
If you don't mind converting your data to tables you can use outerjoin:
A = [1981 0.79 1.56 0.90 1.15;
1982 0.62 0.83 0.84 0.74;
1983 0.81 0.71 0.71 0.70;
1984 1.06 0.74 0.61 0.76;
1985 1.23 0.86 0.67 0.61;
1986 1.32 0.56 1.11 0.76;
1987 0.75 1.06 0.56 1.15;
1988 1.76 1.09 0.88 0.67;
1989 0.90 0.77 0.94 0.77;
1990 0.52 0.52 1.15 0.88];
B = [1981 1.0617;
1982 1.0682;
1985 1.0149;
1986 0.6607;
1987 0.5642;
1988 0.6194;
1989 0.6693;
1990 0.6966];
At = array2table(A);
Bt = array2table(B);
C = outerjoin(At,Bt,'Keys',1,'RightVariables',2);
C = C(:,[1 6 2:5])
C =
10×6 table
A1 B2 A2 A3 A4 A5
____ ______ ____ ____ ____ ____
1981 1.0617 0.79 1.56 0.9 1.15
1982 1.0682 0.62 0.83 0.84 0.74
1983 NaN 0.81 0.71 0.71 0.7
1984 NaN 1.06 0.74 0.61 0.76
1985 1.0149 1.23 0.86 0.67 0.61
1986 0.6607 1.32 0.56 1.11 0.76
1987 0.5642 0.75 1.06 0.56 1.15
1988 0.6194 1.76 1.09 0.88 0.67
1989 0.6693 0.9 0.77 0.94 0.77
1990 0.6966 0.52 0.52 1.15 0.88
  1 个评论
Neuropragmatist
Neuropragmatist 2019-8-9
Or using indexing:
A = [1981 0.79 1.56 0.90 1.15;
1982 0.62 0.83 0.84 0.74;
1983 0.81 0.71 0.71 0.70;
1984 1.06 0.74 0.61 0.76;
1985 1.23 0.86 0.67 0.61;
1986 1.32 0.56 1.11 0.76;
1987 0.75 1.06 0.56 1.15;
1988 1.76 1.09 0.88 0.67;
1989 0.90 0.77 0.94 0.77;
1990 0.52 0.52 1.15 0.88];
B = [1981 1.0617;
1982 1.0682;
1985 1.0149;
1986 0.6607;
1987 0.5642;
1988 0.6194;
1989 0.6693;
1990 0.6966];
A = [A NaN(size(A(:,1)))];
[~,LOCB] = ismember(B(:,1),A(:,1));
A(LOCB(LOCB>0),6) = B(LOCB>0,2);
A = A(:,[1 6 2 3 4 5])
A =
Columns 1 through 4
1981 1.0617 0.79 1.56
1982 1.0682 0.62 0.83
1983 NaN 0.81 0.71
1984 NaN 1.06 0.74
1985 1.0149 1.23 0.86
1986 0.6607 1.32 0.56
1987 0.5642 0.75 1.06
1988 0.6194 1.76 1.09
1989 0.6693 0.9 0.77
1990 0.6966 0.52 0.52
Columns 5 through 6
0.9 1.15
0.84 0.74
0.71 0.7
0.61 0.76
0.67 0.61
1.11 0.76
0.56 1.15
0.88 0.67
0.94 0.77
1.15 0.88

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by