合并消息运行#

langchain_core.messages.utils.merge_message_runs(messages: Sequence[MessageLikeRepresentation] | None = None, **kwargs: Any) → list[BaseMessage] | Runnable[Sequence[MessageLikeRepresentation], list[BaseMessage]][source]#

合并相同类型的连续消息。

注意: ToolMessages 不会被合并，因为每个都有一个独特的工具调用 ID，无法合并。

Args:
消息：要合并的类似消息对象的序列。 chunk_separator：指定在消息块之间插入的字符串。默认是“

“.

Returns:

将连续运行的消息类型合并为单个消息的BaseMessages列表。默认情况下，如果两个要合并的消息都有字符串内容，则合并后的内容是这两个字符串的串联，中间用换行符分隔。可以通过指定chunk_separator来控制插入在消息块之间的分隔符。如果至少有一个消息具有内容块列表，则合并后的内容是内容块列表。

Example:

from langchain_core.messages import (
    merge_message_runs,
    AIMessage,
    HumanMessage,
    SystemMessage,
    ToolCall,
)

messages = [
    SystemMessage("you're a good assistant."),
    HumanMessage("what's your favorite color", id="foo",),
    HumanMessage("wait your favorite food", id="bar",),
    AIMessage(
        "my favorite colo",
        tool_calls=[ToolCall(name="blah_tool", args={"x": 2}, id="123", type="tool_call")],
        id="baz",
    ),
    AIMessage(
        [{"type": "text", "text": "my favorite dish is lasagna"}],
        tool_calls=[ToolCall(name="blah_tool", args={"x": -10}, id="456", type="tool_call")],
        id="blur",
    ),
]

merge_message_runs(messages)

[
    SystemMessage("you're a good assistant."),
    HumanMessage("what's your favorite color\nwait your favorite food", id="foo",),
    AIMessage(
        [
            "my favorite colo",
            {"type": "text", "text": "my favorite dish is lasagna"}
        ],
        tool_calls=[
            ToolCall({"name": "blah_tool", "args": {"x": 2}, "id": "123", "type": "tool_call"}),
            ToolCall({"name": "blah_tool", "args": {"x": -10}, "id": "456", "type": "tool_call"})
        ]
        id="baz"
    ),
]

Parameters:

messages (Union[Sequence[MessageLikeRepresentation], None])
kwargs (Any)

Return type:

联合[列表[BaseMessage], Runnable[序列[MessageLikeRepresentation], 列表[BaseMessage]]]

使用 merge_message_runs 的示例

如何合并相同类型的连续消息