All intercepts are 0 except one? Possible recoding problem

Karen Boyd

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Jul 27, 2018, 5:04:09 PM7/27/18

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Hi!

I've encountered some weird behavior and haven't been able to straighten out what's going on.

This is a sample of the output of summary(fit1,fit.measures=TRUE, standardized=TRUE)

Intercepts: Estimate Std.Err z-value P(>|z|) Std.lv Std.all

.item1 0.000 0.000 0.000

.item2 0.000 0.000 0.000

.item3 0.000 0.000 0.000

.item4 0.000 0.000 0.000

.item5 0.000 0.000 0.000

.item6 0.000 0.000 0.000

.item7 1.170 0.004 298.045 0.000 1.170 3.116

.item8 0.000 0.000 0.000

Item 7 is a variable I attempted to recode.

When solving a different issue, I stumbled across a readout that indicates that lavaan believes this variable has the number of levels it had before recoding, not after. (This is a table that returns after a cfa fails due to items with no variance. It has the following column names: "name idx nobs type exo user mean var nlev lnam"). The nlev and lnam values for item 7 reflect the number of levels and "names" before the recode. I don't know, but I assume this is the cause of the anomaly in the "intercepts" readout, because the only difference between item7 and the other items is that i recoded it.

I have tried to put in some recoded some binary variables as well (hoping that 0,1 instead of 1, NA is the solution to "no variance") and lavaan reports their old number of levels as well.

I have inspected each, and the recoding was successful-- the nlev and lnam values should be different. However, levels() of any item returns NULL, because of their data types, so I can't directly change the levels or names.

Does anyone know what I've done wrong, or how I can fix this?

Thanks,

Karen

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Jeremy Miles

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Jul 27, 2018, 5:07:35 PM7/27/18

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I think we need to see your code, and perhaps something like the result of summary(mydataframe).

Jeremy

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Karen Boyd

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Jul 27, 2018, 6:50:35 PM7/27/18

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I'm afraid it's quite a large model and there are a ton of variables (i.e demographics) in the df that are necessary for other tests, but aren't used in this model. I'll try to pull out what I think is important, but feel free to ask for more.

First, I stored each of the columns into a variable:

i.e.

item1 <- data$item1

item2 <- data$item2

these are all likert style items with, for example, responses of 1, 2, 3, 4, 5 or NA. Some are smaller (ie. 1, 2, 3 or NA).

there are a couple binary ones, too, which I stored in this way:

item3<- data$item3

item3[is.na(item3)] <- 0

I did this because if all the responses are 1 or NA, there's no variance and the model will not run. (I can't confirm that this works, because lavaan doesn't recognize the new level. Therefore I have yet to run the model successfully with these binary variables in it.)

For the anomalous variable described in my initial question, I recoded the variable in order to collapse categories 2 and 3 ("no" and "i don't know")

item7 <- data$item7

item7[item7 == 3] <- 2

item7[item7 == 2] <- 0

Then, I created a model, i.e:

CModel <-

' LV1 =~ item1 + item2 +item3 + item 4

LV2 =~ item5 + item6 + item7

LV3 =~ item8 + item9

LV4=~ item10 + item11 +item12

LV2~ LV3

LV1 ~ LV3

LV4 ~ LV1 + LV2'

(The above mimics the exact structure of the model, but does not include all the items. There are a lot).

Then I fit the model using CFA,

fit1 <- cfa(CModel1, data = data, meanstructure = TRUE,

ordered =c(item1, item2, item4, item5, item6, item7. .. )

When I try to included the binary variables in the model and in ordered(), it fails. If I include the binary variables in the model, but exclude them from ordered(), that at least allows the model to run, but the binary variables are not included in the summary report and the output looks the same as if they were not included in the model.

I have tried including item7 in ordered() and excluding it. When it is included, I get this error:

Error in lav_samplestats_step1(Y = Data, ov.names = ov.names, ov.types = ov.types, :
lavaan ERROR: some categories of variable `CAppHasPolicy' are empty in group 1; frequencies are [181 37 0]

If I keep it excluded, the model converges with the following warning :

Warning message:

In muthen1984(Data = X[[g]], ov.names = ov.names[[g]], ov.types = ov.types, :
lavaan WARNING: trouble constructing W matrix; used generalized inverse for A11 submatrix

I can still get a summary of the model:

summary(fit1,fit.measures=TRUE, standardized=TRUE)

The summary has the problem in the original email, quoted here:

Intercepts: Estimate Std.Err z-value P(>|z|) Std.lv Std.all

.item1 0.000 0.000 0.000

.item2 0.000 0.000 0.000

.item3 0.000 0.000 0.000

.item4 0.000 0.000 0.000

.item5 0.000 0.000 0.000

.item6 0.000 0.000 0.000

.item7 1.170 0.004 298.045 0.000 1.170 3.116

.item8 0.000 0.000 0.000

Other features are a "used" value lower than the "total" (I believe this is due to NA values)

the variances table looks similar to the Intercepts table, in that item 7 is anomalous:

Estimate Std.Err z-value P(>|z|) Std.lv Std.all

.item5 0.773 0.773 0.773

.item6 0.938 0.938 0.938

.item7 0.112 0.010 11.097 0.000 0.112 0.792

.item8 0.477 0.477 0.477

.item9 0.643 0.643 0.643

For the dataframe summary you requested, here's an idea of what the normal, problem, and binary variables look like. Item 7 is the problem variable, item 8 is a very similar variable for which there is no problem and item 3 is a binary variable (which, as you can see, is not recognized as recoded, even though printing item3

item7 item8

Min. :1.000 Min. :1.000

1st Qu.:1.000 1st Qu.:1.000

Median :1.000 Median :1.000

Mean :1.335 Mean :1.377

3rd Qu.:1.000 3rd Qu.:2.000

item3

Min. :1

1st Qu.: 1

Median :1

Mean :1

3rd Qu.:1

item3 looks like this:

[1] 0 0 1 1 0 1 0 0 0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 . . .

item7

[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 . . . NA. . .

item8

[1] 1 1 1 3 1 1 1 1 1 1 1 1 3 1 1 3 1 3 1 1 1 1 2 1 1 2 1 1 2 1 2 1 1 1 . . . NA. . .

(Item 8 did not require recoding to collapse categories 2 and 3)

Let me know if I can provide any more info-- I tried to be concise as I could because I know this is long.

Thanks!
Karen

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Jeremy Miles

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Jul 28, 2018, 1:19:02 AM7/28/18

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On Sat, 28 Jul 2018 at 00:50 Karen Boyd <karen...@gmail.com> wrote:

I'm afraid it's quite a large model and there are a ton of variables (i.e demographics) in the df that are necessary for other tests, but aren't used in this model. I'll try to pull out what I think is important, but feel free to ask for more.

First, I stored each of the columns into a variable:
i.e.

item1 <- data$item1
item2 <- data$item2

these are all likert style items with, for example, responses of 1, 2, 3, 4, 5 or NA. Some are smaller (ie. 1, 2, 3 or NA).

there are a couple binary ones, too, which I stored in this way:
item3<- data$item3
item3[is.na(item3)] <- 0

I did this because if all the responses are 1 or NA, there's no variance and the model will not run. (I can't confirm that this works, because lavaan doesn't recognize the new level. Therefore I have yet to run the model successfully with these binary variables in it.)

For the anomalous variable described in my initial question, I recoded the variable in order to collapse categories 2 and 3 ("no" and "i don't know")

item7 <- data$item7
item7[item7 == 3] <- 2
item7[item7 == 2] <- 0

[snip]

ms. There are a lot).
Then I fit the model using CFA,

fit1 <- cfa(CModel1, data = data, meanstructure = TRUE,
ordered =c(item1, item2, item4, item5, item6, item7. .. )

If that's what you wrote, then it looks like you have pulled the variables out of the data frame, and changed them. But then you ask Lavaan to read the data from the data frame. The data in the data frame are not changed.

What happens if you do:

data$item7[data$item7 == 3] <- 2

data$item7[data$item7 == 2] <- 0

?

Karen Boyd

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Jul 31, 2018, 4:42:47 PM7/31/18

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Oh goodness. Thank you. That explains it and seems obvious now.

I am hesitant to change the data itself in case those "I don't knows" are theoretically valuable and I missed it, but I will create a new column in the data frame and recode there.

Thank you very much!
Cheers,
Karen

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