R: How do I retain the time index when running na.spline() on a timeSeries object? -

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लापता मूल्यों के साथ एक टाइमशरीर ऑब्जेक्ट बनाते समय और na.spline () कार्य निष्पादन एक डेटा संरचना बन जाता है जिसके पास कोई समय इंडेक्स नहीं है। यह कैसे ठीक किया जा सकता है?

एक टाइम शरिअस ऑब्जेक्ट मान लें जैसे कि: 

  # कुछ डेटा सेटअप लाइन्स। X & lt; - '' 1987-01-01 "7.1 एनए 3 "1987-01-02" 5.2 5 2 "1987-01-06" 2.3 एनए 9 'लाइनें।' & Lt; - '"1987-01-01" 55.3 66 45 "1987-01-03" 77.3 87 34' # टाइमशरीज ऑब्जेक्ट बनाओ x & lt; - read.zoo (text = Lines.x) y & lt; - read.zoo (text = lines.y) z & lt; - as.timeSeries (मर्ज (x, y))

z एक उचित समय है शृंखला ऑब्जेक्ट: 

  str (z) समय श्रृंखला: नाम: ऑब्जेक्ट डेटा मैट्रिक्स: आयाम: 4 6 कॉलम नाम: V2.x V3। एक्स V4.x V2.y V3.y V4.y पंक्ति नाम: 1987-01-01 ... 1987-01-06 स्थिति: शुरू: 1987-01-01 समाप्ति: 1987-01-06 साथ: प्रारूप:% Y-% m-% d FinCenter: GMT इकाइयां: V2.x V3.x V4.x V2.y V3.y V4.y शीर्षक: टाइम सीरीज़ ऑब्जेक्ट डॉक्युमेंटेशन: मंगल 4 नवंबर 19:09:43 2014 सिर (जेड) GMT   V2.x V3.x V4.x V2.y V3.y V4.y 1987-01-01 7.1 एनए 3 55.3 66 45 1987-01-02 5.2 5 2 एनए एनए 1987-01-03 एनए एनए 77.3 87 34 1987-01-06 2.3 एनए 9 एनए एनए

लेकिन जब मैं Q & lt; - na.spline (z) चलाता हूं, परिणाम में समय गुम है जानकारी। 

  str (Q) num [1: 4, 1: 6] 7.1 5.2 3.6 2.3 5 5 5 5 3 2 ... - attr (*, "dimnames") = सूची 2 का .. $: NULL .. $: chr [1: 6] "V2.x" "V3.x" "V4.x" "V2.y" ... सिर (क्यू) V2.x V3.x V4.x V2.y V3.y V4.y [1,] 7.1 5 3 55.3 66.0 45.0 [2,] 5.2 5 2 66.3 76.5 39.5 [3,] 3.6 5 4 77.3 87.0 34.0 [4,] 2.3 5 9 88.3 97.5 28.5

टाइम इंडेक्स की हानि या परेशानियों से कैसे बचा जा सकता है?

संपादित करें: आगे उत्तेजना जोड़ने के लिए, कमांड के निम्न अनुक्रम कम से कम कुछ समय एक चिड़ियाघर वस्तु में सूचकांक (लेकिन केवल तभी जब as.zoo () बाहरी रूपांतरण होता है): 

  प्रश्न & lt; - as.zoo (as.timeSeries ( Z)) क्यू & lt; - na.splin ई (क्यू)


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