将数据帧分成相等的部分并存储结果
split a dataframe into equal parts and store the results
我对 R 比较陌生。
我有一个大数据框，我想围绕不同的值分成多个数据框。
structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = c("Bangladesh", "Barbados", "Benin", "Burundi",
"Cameroon", "Chile", "Cyprus", "Ecuador", "Equatorial Guinea",
"Gabon", "Ghana", "Guatemala", "Guinea", "Guyana", "Haiti", "India",
"Jordan", "Lebanon", "Liberia", "Madagascar", "Mali", "Mexico",
"Morocco", "Mozambique", "Nepal", "Nicaragua", "Niger", "Oman",
"Pakistan", "Panama", "Peru", "Rwanda", "Senegal", "Seychelles",
"Sierra Leone", "Singapore", "Sri Lanka", "Sudan", "Togo", "Tunisia",
"Turkey", "Uganda", "Zambia"), class = c("pseries", "factor")),
date = structure(12:36, .Label = c("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", "2014", "2015",
"2016", "2017", "2018"), class = c("pseries", "factor")),
oda_gdp = c(0.15080885502447, 0.1744123099839, 0.199176897551553,
0.193616875061556, 0.186942991013889, 0.164744452026834,
0.192609744294439, 0.13752013069625, 0.156186721262664, 0.137192335225767,
0.131167382827501, 0.139945790928319, 0.112553104508006,
0.1172188903714, 0.120991133274215, 0.0940867931618562, 0.0857724612850372,
0.0653099752359248, 0.0714189688493898, 0.0470115150264598,
0.0446068588203229, 0.0414522297087586, 0.0450866627292532,
0.0435203084091358, 0.0404623996092304), entry = c(0, 0,
1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0)), row.names = c(NA, 25L), class = "data.frame")
我想围绕 entry==1 的点将其拆分为多个数据帧。更具体地说，我希望每个新数据框在 entry==1 之前包含 2 行，在 entry==1 之后包含 7 行。
输出应如下所示
df1=structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("Bangladesh", "Barbados", "Benin",
"Burundi", "Cameroon", "Chile", "Cyprus", "Ecuador", "Equatorial Guinea",
"Gabon", "Ghana", "Guatemala", "Guinea", "Guyana", "Haiti", "India",
"Jordan", "Lebanon", "Liberia", "Madagascar", "Mali", "Mexico",
"Morocco", "Mozambique", "Nepal", "Nicaragua", "Niger", "Oman",
"Pakistan", "Panama", "Peru", "Rwanda", "Senegal", "Seychelles",
"Sierra Leone", "Singapore", "Sri Lanka", "Sudan", "Togo", "Tunisia",
"Turkey", "Uganda", "Zambia"), class = c("pseries", "factor")),
date = structure(12:21, .Label = c("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", "2014", "2015",
"2016", "2017", "2018"), class = c("pseries", "factor")),
oda_gdp = c(0.15080885502447, 0.1744123099839, 0.199176897551553,
0.193616875061556, 0.186942991013889, 0.164744452026834,
0.192609744294439, 0.13752013069625, 0.156186721262664, 0.137192335225767
), entry = c(0, 0, 1, 0, 0, 0, 1, 0, 0, 0)), row.names = c(NA,
10L), class = "data.frame")
对于entry=1的第一种情况
df2=structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("Bangladesh", "Barbados", "Benin",
"Burundi", "Cameroon", "Chile", "Cyprus", "Ecuador", "Equatorial Guinea",
"Gabon", "Ghana", "Guatemala", "Guinea", "Guyana", "Haiti", "India",
"Jordan", "Lebanon", "Liberia", "Madagascar", "Mali", "Mexico",
"Morocco", "Mozambique", "Nepal", "Nicaragua", "Niger", "Oman",
"Pakistan", "Panama", "Peru", "Rwanda", "Senegal", "Seychelles",
"Sierra Leone", "Singapore", "Sri Lanka", "Sudan", "Togo", "Tunisia",
"Turkey", "Uganda", "Zambia"), class = c("pseries", "factor")),
date = structure(16:25, .Label = c("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", "2014", "2015",
"2016", "2017", "2018"), class = c("pseries", "factor")),
oda_gdp = c(0.186942991013889, 0.164744452026834, 0.192609744294439,
0.13752013069625, 0.156186721262664, 0.137192335225767, 0.131167382827501,
0.139945790928319, 0.112553104508006, 0.1172188903714), entry = c(0,
0, 1, 0, 0, 0, 0, 0, 0, 0)), row.names = 5:14, class = "data.frame")
对于第二种情况，其中entry=1
我猜 split 可能是可行的方法，但我想不出包含条件的方法。
请注意，所有新创建的数据框都应具有相同的大小。
非常感谢任何帮助！
我认为 split 不是合适的工具。相反，您可以这样做：
lapply(which(df$entry == 1), function(i) df[(i-2):(i+7),])
#> [[1]]
#> country date oda_gdp entry
#> 1 Bangladesh 1976 0.1508089 0
#> 2 Bangladesh 1977 0.1744123 0
#> 3 Bangladesh 1978 0.1991769 1
#> 4 Bangladesh 1979 0.1936169 0
#> 5 Bangladesh 1980 0.1869430 0
#> 6 Bangladesh 1981 0.1647445 0
#> 7 Bangladesh 1982 0.1926097 1
#> 8 Bangladesh 1983 0.1375201 0
#> 9 Bangladesh 1984 0.1561867 0
#> 10 Bangladesh 1985 0.1371923 0
#>
#> [[2]]
#> country date oda_gdp entry
#> 5 Bangladesh 1980 0.1869430 0
#> 6 Bangladesh 1981 0.1647445 0
#> 7 Bangladesh 1982 0.1926097 1
#> 8 Bangladesh 1983 0.1375201 0
#> 9 Bangladesh 1984 0.1561867 0
#> 10 Bangladesh 1985 0.1371923 0
#> 11 Bangladesh 1986 0.1311674 0
#> 12 Bangladesh 1987 0.1399458 0
#> 13 Bangladesh 1988 0.1125531 0
#> 14 Bangladesh 1989 0.1172189 0
#>
#> [[3]]
#> country date oda_gdp entry
#> 13 Bangladesh 1988 0.11255310 0
#> 14 Bangladesh 1989 0.11721889 0
#> 15 Bangladesh 1990 0.12099113 1
#> 16 Bangladesh 1991 0.09408679 0
#> 17 Bangladesh 1992 0.08577246 0
#> 18 Bangladesh 1993 0.06530998 0
#> 19 Bangladesh 1994 0.07141897 0
#> 20 Bangladesh 1995 0.04701152 0
#> 21 Bangladesh 1996 0.04460686 0
#> 22 Bangladesh 1997 0.04145223 0
由 reprex package (v2.0.1)
于 2022-02-08 创建
我对 R 比较陌生。 我有一个大数据框，我想围绕不同的值分成多个数据框。
structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = c("Bangladesh", "Barbados", "Benin", "Burundi",
"Cameroon", "Chile", "Cyprus", "Ecuador", "Equatorial Guinea",
"Gabon", "Ghana", "Guatemala", "Guinea", "Guyana", "Haiti", "India",
"Jordan", "Lebanon", "Liberia", "Madagascar", "Mali", "Mexico",
"Morocco", "Mozambique", "Nepal", "Nicaragua", "Niger", "Oman",
"Pakistan", "Panama", "Peru", "Rwanda", "Senegal", "Seychelles",
"Sierra Leone", "Singapore", "Sri Lanka", "Sudan", "Togo", "Tunisia",
"Turkey", "Uganda", "Zambia"), class = c("pseries", "factor")),
date = structure(12:36, .Label = c("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", "2014", "2015",
"2016", "2017", "2018"), class = c("pseries", "factor")),
oda_gdp = c(0.15080885502447, 0.1744123099839, 0.199176897551553,
0.193616875061556, 0.186942991013889, 0.164744452026834,
0.192609744294439, 0.13752013069625, 0.156186721262664, 0.137192335225767,
0.131167382827501, 0.139945790928319, 0.112553104508006,
0.1172188903714, 0.120991133274215, 0.0940867931618562, 0.0857724612850372,
0.0653099752359248, 0.0714189688493898, 0.0470115150264598,
0.0446068588203229, 0.0414522297087586, 0.0450866627292532,
0.0435203084091358, 0.0404623996092304), entry = c(0, 0,
1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0)), row.names = c(NA, 25L), class = "data.frame")
我想围绕 entry==1 的点将其拆分为多个数据帧。更具体地说，我希望每个新数据框在 entry==1 之前包含 2 行，在 entry==1 之后包含 7 行。
输出应如下所示
df1=structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("Bangladesh", "Barbados", "Benin",
"Burundi", "Cameroon", "Chile", "Cyprus", "Ecuador", "Equatorial Guinea",
"Gabon", "Ghana", "Guatemala", "Guinea", "Guyana", "Haiti", "India",
"Jordan", "Lebanon", "Liberia", "Madagascar", "Mali", "Mexico",
"Morocco", "Mozambique", "Nepal", "Nicaragua", "Niger", "Oman",
"Pakistan", "Panama", "Peru", "Rwanda", "Senegal", "Seychelles",
"Sierra Leone", "Singapore", "Sri Lanka", "Sudan", "Togo", "Tunisia",
"Turkey", "Uganda", "Zambia"), class = c("pseries", "factor")),
date = structure(12:21, .Label = c("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", "2014", "2015",
"2016", "2017", "2018"), class = c("pseries", "factor")),
oda_gdp = c(0.15080885502447, 0.1744123099839, 0.199176897551553,
0.193616875061556, 0.186942991013889, 0.164744452026834,
0.192609744294439, 0.13752013069625, 0.156186721262664, 0.137192335225767
), entry = c(0, 0, 1, 0, 0, 0, 1, 0, 0, 0)), row.names = c(NA,
10L), class = "data.frame")
对于entry=1的第一种情况
df2=structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("Bangladesh", "Barbados", "Benin",
"Burundi", "Cameroon", "Chile", "Cyprus", "Ecuador", "Equatorial Guinea",
"Gabon", "Ghana", "Guatemala", "Guinea", "Guyana", "Haiti", "India",
"Jordan", "Lebanon", "Liberia", "Madagascar", "Mali", "Mexico",
"Morocco", "Mozambique", "Nepal", "Nicaragua", "Niger", "Oman",
"Pakistan", "Panama", "Peru", "Rwanda", "Senegal", "Seychelles",
"Sierra Leone", "Singapore", "Sri Lanka", "Sudan", "Togo", "Tunisia",
"Turkey", "Uganda", "Zambia"), class = c("pseries", "factor")),
date = structure(16:25, .Label = c("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", "2014", "2015",
"2016", "2017", "2018"), class = c("pseries", "factor")),
oda_gdp = c(0.186942991013889, 0.164744452026834, 0.192609744294439,
0.13752013069625, 0.156186721262664, 0.137192335225767, 0.131167382827501,
0.139945790928319, 0.112553104508006, 0.1172188903714), entry = c(0,
0, 1, 0, 0, 0, 0, 0, 0, 0)), row.names = 5:14, class = "data.frame")
对于第二种情况，其中entry=1
我猜 split 可能是可行的方法，但我想不出包含条件的方法。
请注意，所有新创建的数据框都应具有相同的大小。
非常感谢任何帮助！
我认为 split 不是合适的工具。相反，您可以这样做：
lapply(which(df$entry == 1), function(i) df[(i-2):(i+7),])
#> [[1]]
#> country date oda_gdp entry
#> 1 Bangladesh 1976 0.1508089 0
#> 2 Bangladesh 1977 0.1744123 0
#> 3 Bangladesh 1978 0.1991769 1
#> 4 Bangladesh 1979 0.1936169 0
#> 5 Bangladesh 1980 0.1869430 0
#> 6 Bangladesh 1981 0.1647445 0
#> 7 Bangladesh 1982 0.1926097 1
#> 8 Bangladesh 1983 0.1375201 0
#> 9 Bangladesh 1984 0.1561867 0
#> 10 Bangladesh 1985 0.1371923 0
#>
#> [[2]]
#> country date oda_gdp entry
#> 5 Bangladesh 1980 0.1869430 0
#> 6 Bangladesh 1981 0.1647445 0
#> 7 Bangladesh 1982 0.1926097 1
#> 8 Bangladesh 1983 0.1375201 0
#> 9 Bangladesh 1984 0.1561867 0
#> 10 Bangladesh 1985 0.1371923 0
#> 11 Bangladesh 1986 0.1311674 0
#> 12 Bangladesh 1987 0.1399458 0
#> 13 Bangladesh 1988 0.1125531 0
#> 14 Bangladesh 1989 0.1172189 0
#>
#> [[3]]
#> country date oda_gdp entry
#> 13 Bangladesh 1988 0.11255310 0
#> 14 Bangladesh 1989 0.11721889 0
#> 15 Bangladesh 1990 0.12099113 1
#> 16 Bangladesh 1991 0.09408679 0
#> 17 Bangladesh 1992 0.08577246 0
#> 18 Bangladesh 1993 0.06530998 0
#> 19 Bangladesh 1994 0.07141897 0
#> 20 Bangladesh 1995 0.04701152 0
#> 21 Bangladesh 1996 0.04460686 0
#> 22 Bangladesh 1997 0.04145223 0
由 reprex package (v2.0.1)
于 2022-02-08 创建