我试图获得两个时间戳列之间的差异,但毫秒数消失了。
如何纠正这个?
from pyspark.sql.functions import unix_timestamp
timeFmt = "yyyy-MM-dd' 'HH:mm:ss.SSS"
data = [
(1, '2018-07-25 17:15:06.39','2018-07-25 17:15:06.377'),
(2,'2018-07-25 11:12:49.317','2018-07-25 11:12:48.883')
]
df = spark.createDataFrame(data, ['ID', 'max_ts','min_ts']).withColumn('diff',F.unix_timestamp('max_ts', format=timeFmt) - F.unix_timestamp('min_ts', format=timeFmt))
df.show(truncate = False)
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假设您已经有一个包含时间戳类型列的数据框:
from datetime import datetime
data = [
(1, datetime(2018, 7, 25, 17, 15, 6, 390000), datetime(2018, 7, 25, 17, 15, 6, 377000)),
(2, datetime(2018, 7, 25, 11, 12, 49, 317000), datetime(2018, 7, 25, 11, 12, 48, 883000))
]
df = spark.createDataFrame(data, ['ID', 'max_ts','min_ts'])
df.printSchema()
# root
# |-- ID: long (nullable = true)
# |-- max_ts: timestamp (nullable = true)
# |-- min_ts: timestamp (nullable = true)
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您可以通过将时间戳类型列转换为某种double类型来获取以秒为单位的时间,或者通过将该结果乘以 1000 来获取以毫秒为单位的时间(long如果需要整数,则可以选择转换为 to )。例如
df.select(
F.col('max_ts').cast('double').alias('time_in_seconds'),
(F.col('max_ts').cast('double') * 1000).cast('long').alias('time_in_milliseconds'),
).toPandas()
# time_in_seconds time_in_milliseconds
# 0 1532538906.390 1532538906390
# 1 1532517169.317 1532517169317
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最后,如果您想要两次时间之间的差异(以毫秒为单位),您可以这样做:
df.select(
((F.col('max_ts').cast('double') - F.col('min_ts').cast('double')) * 1000).cast('long').alias('diff_in_milliseconds'),
).toPandas()
# diff_in_milliseconds
# 0 13
# 1 434
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我正在 PySpark 2.4.2 上执行此操作。无需使用任何字符串连接。
这是预期的行为- 它在源代码文档字符串unix_timestamp中明确指出它只返回秒,因此在进行计算时会删除毫秒部分。
如果您想进行该计算,可以使用该substring函数连接数字,然后进行差分。请参阅下面的示例。请注意,这假设数据完整,例如毫秒完全满足(所有 3 位数字):
import pyspark.sql.functions as F
timeFmt = "yyyy-MM-dd' 'HH:mm:ss.SSS"
data = [
(1, '2018-07-25 17:15:06.390', '2018-07-25 17:15:06.377'), # note the '390'
(2, '2018-07-25 11:12:49.317', '2018-07-25 11:12:48.883')
]
df = spark.createDataFrame(data, ['ID', 'max_ts', 'min_ts'])\
.withColumn('max_milli', F.unix_timestamp('max_ts', format=timeFmt) + F.substring('max_ts', -3, 3).cast('float')/1000)\
.withColumn('min_milli', F.unix_timestamp('min_ts', format=timeFmt) + F.substring('min_ts', -3, 3).cast('float')/1000)\
.withColumn('diff', (F.col('max_milli') - F.col('min_milli')).cast('float') * 1000)
df.show(truncate=False)
+---+-----------------------+-----------------------+----------------+----------------+---------+
|ID |max_ts |min_ts |max_milli |min_milli |diff |
+---+-----------------------+-----------------------+----------------+----------------+---------+
|1 |2018-07-25 17:15:06.390|2018-07-25 17:15:06.377|1.53255330639E9 |1.532553306377E9|13.000011|
|2 |2018-07-25 11:12:49.317|2018-07-25 11:12:48.883|1.532531569317E9|1.532531568883E9|434.0 |
+---+-----------------------+-----------------------+----------------+----------------+---------+
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