In scanning through a simple imported file of weather station data, some dates have missing data.
What is a test to use to check whether a datum is valid? I tried If[tempData[index][[4]]≠NaN, ]
and If[tempData[index][[4]]≠Indeterminate, ]
, but neither seemed to work.
Most of my data is valid, and so it is just a matter of skipping/ignoring the occasional missing entry, and not incrementing my accumulators or counters.
The data are already arranged in a Mathematica list, and a missing entry is represented by two adjacent commas with nothing in between.
So, what I need is an If[ ]
statement that will trigger (or not trigger) when it encounters such a missing data element.
Here is the code, along with a first snippet of output:
f[x_ /; MemberQ[Range@12, x]] :=
Switch[x, 2, 28, 4 | 6 | 9 | 11, 30, _, 31]
Do[
Do[ sum = 0.0; count = 0; index = 1;
While[index < 20454,
If[Part[tempData, index][[{2, 3}]] == {month, day},
sum = sum + Part[tempData, index][[4]]; count++ ]; index++];
Print[{month, day}, " ", sum, " ", count],
{day, 1, f[month]}
],
{month, 1, 12}
]
{1,1} 87.5 +5 56
{1,2} 60.4 +5 56
{1,3} 59.5 +5 56
{1,4} 54.5 +5 56
{1,5} 37.4 +5 56
{1,6} 39.6 +5 56
I interpret these results to mean that Mathematica found, for these six days in January, the sums of the temperatures for the corresponding days in the 20454 days of data (about 56 years) contained in tempData which had valid data. I think the +5 means that the program encountered 5 Null values, so that the data count should really be 51. If I wanted an average temperature for 01 January based on these data, I would divide 87.5 by 51.
My main concern is not computing the average, or any of the other manipulations I plan to do, but merely how to skip the Null data values. My most recent attempt was to add
If[!Null === Part[tempData, index][[4]], sum = sum + Part[tempData, index][[4]], count++]
but that did worse than not work — the sums and counts are all zero.
Here is a sample of the input data, tempData:
{{1955, 1, 1, 0.4, 1.9, -0.3, 4.2},
{1955, 1, 2, -0.5, 0.5, -1.1, 2.5},
{1955, 1, 3, 0.5, 2.7, -1.9, 0.8},
{1955, 1, 4, 1.4, 2.6, 0.6,"0T"},
{1955, 1, 5, 0.8, 2.3, -0.3, 0},
{1955, 1, 6, 0.7, 3.7, -1.7, 0}, etc. }
Null
, I almost pulled all of my hair out.value = Null; If[Null == value, "T", "F", "O"]
yieldsT
as expected, butvalue = 1; If[Null == value, "T", "F", "O"]
yieldsO
notF
as one might think. The trick is when testing forNull
to use three=
's.value = Null; If[Null === value, "T", "F", "O"]
yieldsT
andvalue = 1; If[Null === value, "T", "F", "O"]
yieldsF
as expected. $\endgroup$ – mmorris Mar 7 '12 at 0:42If[!Null === value,
$\endgroup$ – mmorris Mar 7 '12 at 17:05