Question

我有一个包含少量列的数据集，但是当列1具有偶数时，我想将其分隔为新文件。第1列格式为Var1 Var2 Var3 M1 * 2 3 M3 * 4 2 M2 * 1 5 M6 * 1 6，此处为奇数（1）。

数据集：

/**
 * This pans the map when the user is moused down and at the end of the map
 * bounding box.
 *
 * @return {undefined}
 */
DrawingManager.prototype._mapPanHandler = function (googleMapEvent) {
  var self = this;
  var event = googleMapEvent;

  var overlay = new google.maps.OverlayView();
  overlay.draw = function() {};
  overlay.setMap(map);

  var projection = overlay.getProjection();
  var pixelpoint = projection.fromLatLngToDivPixel(map.getCenter());

  var proj = overlay.getProjection();
  var thepix = proj.fromLatLngToDivPixel(event.latLng);

  var nE = map.getBounds().getNorthEast();
  var sW = map.getBounds().getSouthWest();
  var nE = proj.fromLatLngToDivPixel(nE);
  var sW = proj.fromLatLngToDivPixel(sW);

  var north = nE.y;
  var east  = nE.x;
  var south = sW.y;
  var west  = sW.x;

  var appx_north, appx_east, appx_south, appx_west;

  if (Math.round(thepix.y) <= Math.round(north + 50)) {appx_north = true;}
  if (Math.round(thepix.x) >= Math.round(east  - 50)) {appx_east  = true;}
  if (Math.round(thepix.y) >= Math.round(south - 50)) {appx_south = true;}
  if (Math.round(thepix.x) <= Math.round(west  + 50)) {appx_west  = true;}

  clearInterval(self._panTimer);

  if (!appx_north && !appx_east && !appx_south && !appx_west) {
    return;
  }

  var panMap = function () {
    var movementAmount = 3;
    if (appx_north) { pixelpoint.y -= movementAmount; }
    if (appx_east)  { pixelpoint.x += movementAmount; }
    if (appx_south) { pixelpoint.y += movementAmount; }
    if (appx_west)  { pixelpoint.x -= movementAmount; }
    point = projection.fromDivPixelToLatLng(pixelpoint);
    map.setCenter(point);
  }

  panMap();
  self._panTimer = setInterval(panMap, 10);
};

Answer 1

试试这个：

$ awk '!(substr($1,2)%2)' infile > new_file
$ cat new_file
Var1 Var2 Var3
M2 * 1 5
M6 * 1 6

Answer 2

使用grep：

class Data{
public:
    int a;
    Data(int A): a(A) {}
    operator int()  {return a;}
};

int operator=(int &lhs, Data* rhs){
    lhs = rhs->a;
    return lhs;
}

int main(){
    Data d(10);
    int A = &d;
    return 0;
}

这将获得第一个空格前的字符均匀的所有行，即第一列以偶数结束。

Answer 3

awk救援！

$ awk '$1~/[02468]$/' file   

M2 * 1 5 
M6 * 1 6

检查第一个字段的最后一个数字是否是其中一个。

添加标题

$ awk 'NR==1 || $1~/[02468]$/' file

Var1 Var2 Var3 
M2 * 1 5 
M6 * 1 6

如何使用awk将文件分隔到不同的文件

3 个答案: