You want to create a LangChain workflow that takes a list of numbers, passes it unchanged, then doubles each number. Which combination of RunnablePassthrough and RunnableLambda is correct?
hard📝 Application Q15 of 15
LangChain - Chains and LCEL
You want to create a LangChain workflow that takes a list of numbers, passes it unchanged, then doubles each number. Which combination of RunnablePassthrough and RunnableLambda is correct?
AUse RunnablePassthrough twice, no lambda needed
BUse RunnableLambda with lambda x: x, then RunnablePassthrough to double numbers
CUse RunnableLambda with lambda x: x*2, then RunnablePassthrough to pass list
DUse RunnablePassthrough to pass the list, then RunnableLambda with lambda x: [i*2 for i in x]
Step-by-Step Solution
Solution:
Step 1: Pass list unchanged with RunnablePassthrough
RunnablePassthrough returns the list as is, so it fits the first step.
Step 2: Double each number with RunnableLambda
RunnableLambda with lambda x: [i*2 for i in x] correctly doubles each element in the list.
Final Answer:
Use RunnablePassthrough to pass the list, then RunnableLambda with lambda x: [i*2 for i in x] -> Option D
Quick Check:
Passthrough passes list, lambda doubles elements = A [OK]
Quick Trick:Passthrough passes input; lambda transforms list elements [OK]
Common Mistakes:
Trying to double list with passthrough
Using lambda that multiplies list object, not elements
Reversing order of passthrough and lambda
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